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C - Assignment makes integer from pointer without a cast
I'm getting this error: Warning: assignment makes integer from pointer without a cast [enabled by default] This is my source code:
So, I'm getting an error ( warning ) and my result of my function ft_replace is fxc... It return me a result like 4227111 and I don't know why I would like my function when it find the last char here this is ' - ' it return the position of the array.
So a string which contains " Hel-l-o " -> Return 6 A string which contains " He-llo " -> Return 3 A string which contains " Hell-o- " -> Return 7
- 2 Well.. it's pretty clearly said - s1 is pointer, not a number(index). You can use additional counter, to count the index/position, or use pointer arithmetic to calculate the position. – Kiril Kirov Commented May 21, 2015 at 13:34
- @user3208656 And what to return if the character is not found in a string? – Vlad from Moscow Commented May 21, 2015 at 13:35
- What is the intent of this function? Are you trying to return the index of the last '-' within s1? If so then you probably want this instead: int ft_replace(char const *s1) { int result; result = 0; for (const char* s = s1; *s != '\0'; ++s) { if (*s == '-') result = (s1-s); } return (result); } – Ron Kuper Commented May 21, 2015 at 13:39
4 Answers 4
result = s1
result is an int , integer s1 is a char const* , a const pointer to char
You can't assign the value of a const char* to an integer (because they might not have the same size for example).
You can however, as hinted by the warning, cast the value to the correct type to force it to stop warning you:
But this is not a good practice as it can induce undefined behavior. Reading your code, I don't think you are doing what you think you are doing anyway when you assign s1 to result . What you should be doing, is incrementing a counter and set result during the loop:
In your code, result is int and s1 is char const * .
You can change it to
However, there is no way to know the position of the array currently. You need to have a separate counter to keep track of the valid element present in the array .
- 3 @Eregrith: You certainly can assign a char to an int , it just won't do anything useful in this case. – Bill Lynch Commented May 21, 2015 at 13:37
- @Eregrith what about §6.5.16.1, paragraph 2? – Natasha Dutta Commented May 21, 2015 at 13:42
- @KirilKirov well, OP has logical issue in his code, that I have pointed out later. Otherwise, as per the question title, this is the way to remove the error. Now whether if fits the logic, that is another aspect (which also I addressed, I think). – Natasha Dutta Commented May 21, 2015 at 13:51
- @BillLynch And NatashaDutta you are both right. My bad – Eregrith Commented May 21, 2015 at 13:51
- @Natasha Dutta How do you do?:) – Vlad from Moscow Commented May 21, 2015 at 13:52
I think you mean the following
As for your code then this statement
does not make sense. In the left side of the assignment there is an object of type int while in the right side of the assignment there is a pointer of type const char *
If you want to return pointer to the last occurence of character '-' then the function can look like
A pointer holds the memory address, not the position in the array.
Use an index variable to keep track of the position.
Note that the first element in an array in C has index 0 , hence the +1 when assigning result :
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C: warning assignment makes integer from pointer without a cast
- Thread starter eatsleep
- Start date Sep 13, 2013
- Tags Assignment Integer
- Sep 13, 2013
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"RED" is not a character variable - it is a string. The pointer conversion is from the string pointer.
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jedishrfu said: choice should be defined as: char *choice;
D H said: That should be const char* choice , not just char* choice . Assigning to a pointer to a string such as "RED" is illegal (undefined behavior), so it's best to make the pointer a type that does not accept assignments.
jedishrfu said: Doesn't this depend on what he's trying to do? Suppose this choice is in some sort of input loop where first its RED and then its BLUE ...
D H said: Assigning to a pointer to a string such as "RED" is illegal (undefined behavior), so it's best to make the pointer a type that does not accept assignments.
D H said: There's nothing wrong with that. const char * choice (or char const * choice , same thing) means that choice can be used on the left hand side of an assignment statement but that choice[1] cannot. You are apparently thinking of char * const choice = "RED"; , but that's a completely different data type. With this declaration, choice can only be assigned a value in the declaration statement. However, choice[1]='D'; is perfectly legal with this declaration. You can combine the two restrictions with the declaration const char * const choice = "RED";
- Sep 14, 2013
rcgldr said: It's my understanding that literal strings are like statics, and exist from start to termination of a program, so why should assigning a pointer to a literal string be undefined? Trying to change a value in the literal string via the pointer would be illegal / undefined behavior, but the pointer assignment shouldn't be an issue.
nsaspook said: It depends on the computer architecture and how strict the compiler is.
rcgldr said: The point I was making is how string literals are defined in the C89 and later standards. From the C89 standard, section 3.1.4: ... So it would seem that only an attempt to modify a string is undefined, not the usage of a pointer to access a literal string. The type "array of char" or "array of wchar_t" would be the same regardless of where the string literal was stored (ROM, RAM, code section of a program, ... ).
D H said: I wasn't clear with my previous post. There's nothing wrong per se with char* ptr; ...; ptr="RED"; ...; ptr="BLUE"; What's wrong is assigning into that pointer: *ptr = 'A'; . What's worse is that most compilers won't report this as an error. You don't find out until runtime. Declaring the variable as a const char* pointer (or char const* , same thing) and assignments such as ptr="RED"; are still legal, but now *ptr = 'A'; becomes a compile-time error.
Related to C: warning assignment makes integer from pointer without a cast
This warning indicates that a pointer value is being assigned to an integer variable without being explicitly converted or casted. This can create unexpected behavior and should be avoided.
In C, pointers are used to store memory addresses of variables or data. Assigning a pointer value to an integer variable can result in loss of information or incorrect memory access, which is why the compiler issues this warning.
To fix this warning, you can use a typecast to explicitly convert the pointer value to an integer before assigning it to the variable. However, make sure that the conversion is appropriate and does not cause any loss of information.
While this warning may not always result in errors, it can lead to unexpected behavior and should be addressed. Ignoring this warning can potentially cause issues in your program, so it is best to fix it.
In some cases, using a different data type for the variable may be a better solution than using a typecast. For example, if the pointer is pointing to a string, you can use a string data type instead of converting it to an integer.
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Makes Pointer From Integer Without a Cast: Fix It Now!
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Our research team has ensured that this will be your best resource on these error messages, and you won’t have to read another article about it again. With that said, launch your code that’s showing the error messages, and let’s fix it for you.
JUMP TO TOPIC
– You Assigned an Integer to a Pointer
– you want to convert an integer to a pointer, – you passed a variable name to the “printf()” function, – you used “struct _io_file” in a wrong way, – you copied a string to an invalid location, – you’re setting a pointer to a different type, – use equal data types during assignment, – ensure the pointer and integer have the same sizes, – pass a “format string” to the “printf()” function, – use a function that returns a pointer to “struct _io_file”, – copy the string to character array or character pointer, – assign pointers of compatible types, why your code made a pointer from an integer without a cast.
Your code made a pointer from an integer without a cast because you assigned an integer to a pointer or you want to convert an integer to a pointer. Other causes include the passing of a variable name to the “printf()” function and using “struct _io_file in the wrong way.
Finally, the following are also possible causes:
- You copied a string to an invalid location
- You’re setting a pointer to a different type
If you assign an integer to a pointer, it will lead to the “ assignment makes pointer from integer without a cast c programming ” error. For example, in the following, the “num_array” variable is an unsigned integer type while “tmp_array” is a pointer.
Later, an error will occur when the “for” loop tries to copy the values of the “num_array” to the “tmp_array” pointer using a variable assignment.
| #include<stdio.h> #include<stdint.h> uint8_t num_array[8] = {0, 1, 2, 3, 4, 5, 6, 7}; void copy_arr_values() { int i; uint8_t *tmp_array[8]; for(i=0; i<8; i++){ tmp_array[i] = num_array[(i+3)%8]; } } int main() { copy_arr_values(); }
|
For example, in the following, the “theta” variable is an integer while “ptr_theta” is a pointer. Both have different sizes, and you can’t convert the integer to a pointer.
| #include<stdio.h> int main() { int theta = 10; int *ptr_theta = (int *)theta; } |
If you pass a variable name to the “printf()” function, that’s when the compiler will throw the “ passing argument 1 of ‘printf’ makes pointer from integer without a cast ” error.
For example, in the following, “num_val” is an integer variable, and the code is trying to print it using the “printf()” function.
| #include <stdio.h> int main() { int num_val = 42; printf(num_val); //Error! return 0; } |
When you assign an integer to a pointer of “struct _io_file”, that’s when you’ll get the “ assignment to file aka struct _io_file from int makes pointer from integer without a cast ” error. For example, in the following code, “FILE” is a “typedef” for “struct _io_file”. This makes it a pointer to “struct _io_file”.
Later, the code assigned it to the integer “alpha,” and this will lead to an error stated below:
| #include <stdio.h> int main(void) { int alpha = 10; FILE *beta = alpha; return 0; } |
Now, in the following, the code is trying to copy a string (“source”) into an integer (“destination”), and this leads to an error because it’s not a valid operation:
| #include<stdio.h> #include<string.h> int main() { int destination; char source[] = “Hello, World!”; strcpy(destination, source); return 0; } |
When your code sets a pointer to a different type, your compiler will show the “ incompatible pointer type ” error. In the following code, “pacifier” is an integer pointer, while “qwerty” is a character pointer.
Both are incompatible, and your C compiler will not allow this or anything similar in your code.
| #include<stdio.h> int main() { int charlie = 5; int *pacifier = &charlie; char *qwerty = pacifier; } |
How To Stop a Pointer Creation From an Integer Without a Cast
You can stop a pointer creation from an integer without a cast if you use equal data types during the assignment or ensure the integer and pointer have the same sizes. What’s more, you can pass a “format string” to “printf()” and use a function that returns a pointer to “struct _io_file”.
- Copy the string to a character array or character pointer
- Assign pointers of compatible types
During a variable assignment, use variables of equal data types. In our first example, we assigned a pointer (uint8_t *tmp_array) to an unsigned integer (uint8_t num_array) and it led to a compile-time error.
Now, the following is the revised code, and we’ve changed “tmp_array” to a real array. This will prevent the “ gcc warning: assignment makes pointer from integer without a cast ” error during compilation.
| #include<stdio.h> #include<stdint.h> uint8_t num_array[8] = {0, 1, 2, 3, 4, 5, 6, 7}; int main() { int i; // Fix: change tmp_array to an array uint8_t tmp_array[8] = {}; for (i = 0; i < 8; i++){ tmp_array[i] = num_array[(i+3)%8]; } int length = sizeof(tmp_array)/sizeof(tmp_array[0]); printf(“Elements of the ‘tmp_array’: \n”); for (int i = 0; i < length; i++) { printf(“%d “, tmp_array[i]); } } |
Now, the fix is to use “intptr_t” from the “stdint.h” header file because it guarantees that the pointer and integer will have the same sizes. We’ve used it in the following code, and you can compile it without an error.
| #include<stdio.h> #include<stdint.h> int main() { int theta = 10; int *ptr_theta = (int *)(intptr_t)theta; printf(“%p”, ptr_theta); } |
To fix the “pointer to integer without a cast” in the “printf()” function, pass a “format string” as its first argument. How you write the “format string” depends on the variable that you’ll print. In the following updated code, “num_val” is an integer, so the “format string” is “%d”.
| #include <stdio.h> int main() { int num_val = 42; printf(“%d”, num_val); return 0; } |
When you’re using “FILE” from the “stdlib.h” header file, you can prevent any “pointer from integer” error if you use a function that returns a pointer to “struct _io_file”.
An example of such a function is “fopen()” which allows you to open a file in C programming. Now, the following is a rewrite of the code that causes the pointer error in “struct _io_file”. This time, we use “fopen()” as a pointer to “FILE”.
| #include <stdio.h> int main(void) { FILE *f = fopen(“myfile.txt”, “r”); if (f == NULL) { perror(“Error opening file”); return 1; } char c; while ((c = fgetc(f)) != EOF) { printf(“%c”, c); } return 0; } |
Now, in the following code, we’ve changed “destination” from an integer to a “character array”. This means “strcpy()” can copy a string into this array without an error.
| #include<stdio.h> #include<string.h> int main() { char destination[100]; char source[] = “Hello, World!”; strcpy(destination, source); printf(“%s”, destination); return 0; } |
Meanwhile, the following is another version that turns the “destination” into a “character pointer”. With this, you’ll need to allocate memory using the “malloc()” function from the “stdlib.h” header file.
| #include<stdio.h> #include<string.h> #include<stdlib.h> int main() { char* destination; char source[] = “Hello, World!”; destination = malloc(100); strcpy(destination, source); printf(“%s”, destination); free(destination); return 0; } |
When you’re assigning pointers, ensure that they’re compatible by changing their data type . The following is the updated code for the “charlie” example , and “qwerty” is now an integer.
| #include<stdio.h> int main() { int charlie = 5; int *pacifier = &charlie; int *qwerty = pacifier; } |
This article explained why your code made a pointer without a cast and how you can fix it. The following is a summary of what we talked about:
- An attempt to convert an integer to a pointer will lead to the “makes pointer from integer without a cast wint conversion” error.
- To prevent an “incompatible pointer type” error, don’t set a pointer to a different type.
- If you’re using the “printf()” function, and you want to prevent any “pointer to integer” error, always pass a “format specifier”.
At this stage, you’ll be confident that you can work with integers and pointers in C programming without an error. Save our article, and share it with your developer communities to help them get rid of this error as well.
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[SOLVED] C - assigment makes integer from pointer without a cast warning
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Thread: [SOLVED] C - assigment makes integer from pointer without a cast warning
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I know it's a common error, and I've tried googling it, looked at a bunch of answers, but I still don't really get what to do in this situation.... Here's the relevant code: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char path[51]; const char* home = getenv( "HOME" ); strcpy( path, argv[1] ); path[1] = home; return 0; } -- there is more code in the blank lines, but the issue's not there (I'm fairly sure), so didn't see the point in writing out 100 odd lines of code. I've tried some stuff like trying to make a pointer to path[1], and make that = home, but haven't managed to make that work (although maybe that's just me doing it wrong as opposed to wrong idea?) Thanks in advance for any help
Re: C - assigment makes integer from pointer without a cast warning
path[1] is the second element of a char array, so it's a char. home is a char *, i.e. a pointer to a char. You get the warning because you try to assign a char* to a char. Note also the potential to overflow your buffer if the content of the argv[1] argument is very long. It's usually better to use strncpy.
Last edited by r-senior; March 10th, 2013 at 03:03 PM . Reason: argv[1] would overflow, not HOME. Corrected to avoid confusion.
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You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that.
Last edited by Christmas; March 10th, 2013 at 09:51 PM .
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Originally Posted by Christmas You can try something like this: Code: #include <stdio.h> #include <stdlib.h> #include <string.h> int main( int argc, char *argv[] ) { char *path; const char *home = getenv("HOME"); path = malloc(strlen(home) + 1); if (!path) { printf("Error\n"); return 0; } strcpy(path, home); printf("path = %s\n", path); // if you want to have argv[1] concatenated with path if (argc >= 2) { path = malloc(strlen(home) + strlen(argv[1]) + 1); strcpy(path, argv[1]); strcat(path, home); printf("%s\n", path); } // if you want an array of strings, each containing path, argv[1]... char **array; int i; array = malloc(argc * sizeof(char*)); array[0] = malloc(strlen(home) + 1); strcpy(array[0], home); printf("array[0] = %s\n", array[0]); for (i = 1; i < argc; i++) { array[i] = malloc(strlen(argv[i]) + 1); strcpy(array[i], argv[i]); printf("array[%d] = %s\n", i, array[i]); } // now array[i] will hold path and all the argv strings return 0; } Just as above, your path[51] is a string while path[1] is only a character, so you can't use strcpy for that. Excellent point. I've basically fixed my problem by reading up on pointers again (haven't done any C for a little while, so forgot some stuff), and doing: Code: path[1] = *home; the code doesn't moan at me when I compile it, and it runs okay (for paths which aren't close to 51 at least), but after reading what you read, I just wrote a quick program and found out that getenv("HOME") is 10 characters long, not 1 like I seem to have assumed, so I'll modify my code to fix that.
Yes, getenv will return the path to your home dir, for example /home/user, but path[1] = *home will still assign the first character of home to path[1] (which would be '/').
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A return type of ctypes.c_void_p is also "helpful" and converts the returned C address to a Python integer, but can be cast a more specific pointer type to access the data at the address. To find it's address, use ctypes.addressof on the contents of the pointer; otherwise you get the address of the storage of the pointer.
The warning comes from the fact that you're dereferencing src in the assignment. The expression *src has type char, which is an integral type.The expression "anotherstring" has type char [14], which in this particular context is implicitly converted to type char *, and its value is the address of the first character in the array.So, you wind up trying to assign a pointer value to an integral ...
1. Earlier, I asked a similar question, but I've since changed my code. Now the compiler gives me a different warning. This is an example of what my code looks like now: void *a = NULL; void *b = //something; a = *(int *)((char *)b + 4); When I try to compile, I get "warning: assignment makes pointer from integer without a cast."
3. result = s1. result is an int, integer. s1 is a char const*, a const pointer to char. You can't assign the value of a const char* to an integer (because they might not have the same size for example). You can however, as hinted by the warning, cast the value to the correct type to force it to stop warning you: result = (int)s1;
Case 1: Assignment of a pointer to an integer variable. n1 = 2; ptr = &n1; n2 = ptr; /* Failure in this line */. In this simple code we have three variables, an integer pointer "ptr", and two ...
Converting Integers to Pointers {#converting-integers-to-pointers} To convert an integer to a pointer, follow these steps: Include the <stdint.h> header (C) or the <cstdint> header (C++) in your program. Cast your integer to the required pointer type using a double cast. int a = 42; uintptr_t int_ptr = (uintptr_t)&a;
OK - in that case, since you want to immediately change the value of sad, it can be a local variable to the function, and doesn't need to be an argument to the function: . void racun(int D, int M, int G, int *dUg) { int i, *sad; If you want sad to point to an array, such that accessing sad (sad[...]) will be equivalent to accessing that array (e.g. obicna[...]), you can simply assign the array ...
text should be declared as: char *text = NULL; You need a pointer to char. This pointer can be set to point to any literal string (as you do in your case statements). char text; // this is just a single character (letter), not a string. 2. Objective_Ad_4587 • 3 yr. ago. i got it thank you very much.
Related to C: warning assignment makes integer from pointer without a cast 1. What does the warning "assignment makes integer from pointer without a cast" mean? This warning indicates that a pointer value is being assigned to an integer variable without being explicitly converted or casted. This can create unexpected behavior and should be avoided.
Understand its causes and solution in this guide. (initialization makes pointer from integer without a cast) ... Check if the variable on the left-hand side of the assignment operator is a pointer variable. If it is not a pointer variable, you need to declare it as a pointer variable.
How To Stop a Pointer Creation From an Integer Without a Cast. - Use Equal Data Types During Assignment. - Ensure the Pointer and Integer Have the Same Sizes. - Pass a "Format String" to the "Printf ()" Function. - Use a Function That Returns a Pointer to "Struct _IO_file". - Copy the String to Character Array or Character ...
Re: C - assigment makes integer from pointer without a cast warning. path [1] is the second element of a char array, so it's a char. home is a char *, i.e. a pointer to a char. You get the warning because you try to assign a char* to a char. Note also the potential to overflow your buffer if the content of the argv [1] argument is very long.
Assignment makes pointer from integer without a cast Getting started with C or C++ | C Tutorial | C++ Tutorial | C and C++ FAQ | Get a compiler | Fixes for common problems Thread: Assignment makes pointer from integer without a cast
char *word is a pointer, meaning the value contained by the word parameter is a memory address (in this example, it will be the address of the first character in a string). When you use the notation *word on line 19, you are dereferencing the word pointer, meaning it will return the value of whatever is stored at word's memory address (which is a single character).
warning: assignment makes pointer from integer without a cast I get that when I compile a C file where the function is defined as this: char **getvars () and the calling function has this declared: char **s; and I write: s=getvars (); C / C++. 1. 3858. warning assignment makes integer from pointer without a cast.
This is my last program, and I have done quite well with the other 15. Errors: |43|warning: assignment makes integer from pointer without a cast [enabled by default]| |44|warning: assignment makes integer from pointer without a cast [enabled by default]| |46|error: incompatible types when assigning to type 'int [100]' from type 'int *'| |47 ...
Solving Assignment Makes Integer from Pointer Without a Cast in C: Tips and Solutions
p = itoa (num); You probably want: p = itoa (*num); Since otherwise you're sending a pointer to an int, and not the int. But I don't think itoa () even works that way. You'd also have to send in the string buffer. You should go check out a reference. 1. I'm new to C language: include "main.h" int func_d (va_list li) { char *p; int *num; int ...
assignment makes integer from pointer without a castc/c++ warning explained#syntax #c/c++ #compiler #error #warning
I'm not sure what I'm doing wrong, but I'm getting "assignment makes integer from pointer without a cast" at each first line of the internal while loops (the ones that only have " {" in them). Anyone have a clue what I'm doing wrong? And I wouldn't use strcmp () to find your comma's. Just.