assignment makes pointer from integer without a cast c programming

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Assignment makes integer from pointer without a cast in c

Programming can be both rewarding and challenging. You work hard on your code, and just when it seems to be functioning perfectly, an error message pops up on your screen, leaving you frustrated and clueless about what went wrong. One common error that programmers encounter is the "Assignment makes integer from pointer without a cast" error in C.

This error occurs when you try to assign a value from a pointer variable to an integer variable without properly casting it. To fix this error, you need to make sure that you cast the pointer value to the appropriate data type before assigning it to an integer variable. In this article, we will dive deeper into the causes of this error and provide you with solutions to overcome it.

What makes this error occur?

I will present some cases that triggers that error to occur, and they are all have the same concept, so if you understanded why the failure happens, then you will figure out how to solve all the cases easily.

Case 1: Assignment of a pointer to an integer variable

In this simple code we have three variables, an integer pointer "ptr" , and two integers "n1" and "n2" . We assign 2 to "n1" , so far so good, then we assign the address of "n2" to "ptr" which is the suitable storing data type for a pointer, so no problems untill now, till we get to this line "n2 = ptr" when we try to assign "ptr" which is a memory address to "n2" that needs to store an integer data type because it's not a pointer.

Case 2: Returning a Pointer from a Function that Should Return an Integer

As you can see, it's another situation but it's the same idea which causes the compilation error.  We are trying here to assign the return of getinteger function which is a pointer that conatains a memory address to the result variable that is an int type which needs an integer

Case 3: Misusing Array Names as Pointers

As we might already know, that the identifier (name) of the array is actually a pointer to the array first element memory address , so it's a pointer after all, and assigning a pointer type to int type causes the same compilation error.

The solutions

The key to avoiding the error is understanding that pointers and integers are different types of variables in C. Pointers hold memory addresses, while integers hold numeric values. We can use either casting , dereferencing the pointer or just redesign another solution for the problem we are working on that allows the two types to be the same. It all depending on the situation.

Let's try to solve the above cases:

Case 1: Solution: Deferencing the pointer

We need in this case to asssign an int type to "n2" not a pointer or memory address, so how do we get the value of the variable that the pointer "ptr" pointing to? We get it by deferencing the pointer , so the code after the fix will be like the following:

Case 2: Solution: Choosing the right data type

In this case we have two options, either we change the getinteger returning type to int or change the result variable type to a pointer . I will go with the latter option, because there are a lot of functions in the C standard library that returning a pointer, so what we can control is our variable that takes the function return. So the code after the fix will be like the following:

We here changed the result variable from normal int to an int pointer by adding "*" .

Case 3: Solution: Using the array subscript operator

In this case we can get the value of any number in the array  by using the subscript opeartor ([]) on the array with the index number like: myarray[1] for the second element which is 2 . If we still remember that the array identifier is a pointer to the array first memory, then we can also get the value of the array first element by deferencing the array identifier like: *myarray which will get us 1 .

But let's solve the case by using the subscript opeartor which is the more obvious way. So the code will be like the following:

Now the number 1 is assigned to myint without any compilation erros.

The conclusion

In conclusion, the error "assignment to ‘int’ from ‘int *’ makes integer from pointer without a cast" arises in C programming when there is an attempt to assign a memory address (held by a pointer) directly to an integer variable. This is a type mismatch as pointers and integers are fundamentally different types of variables in C.

To avoid or correct this error, programmers need to ensure they are handling pointers and integers appropriately. If the intent is to assign the value pointed by a pointer to an integer, dereferencing should be used. If a function is meant to return an integer, it should not return a pointer. When dealing with arrays, remember that the array name behaves like a pointer to the first element, not an individual element of the array.

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[SOLVED] C - assigment makes integer from pointer without a cast warning

Thread: [solved] c - assigment makes integer from pointer without a cast warning, thread tools.

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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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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.

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.

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.

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:

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:

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.

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.

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.

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”.

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”.

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.

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.

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.

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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assignment makes pointer from integer without a cast

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Thread: assignment makes pointer from integer without a cast

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Hello, I'm hoping someone can explain this warning to me. I'm sure it's just something I'm not aware of! Code: if ((infile = fopen("/src/infile", "r")) == NULL) { printf("Couldn't Open File"); exit(1); } else { /*READ CHARS INTO THE MM STORAGE*/ char c; char d; do{ c = fgetc(infile); //printf("%c\n", c); if(c != '\n') { d = fgetc(infile); MM[I_COUNTER] = c; //<---Warning Here I_COUNTER++; MM[I_COUNTER] = d; //<---Warning Here I_COUNTER ++; } } while((c != 'D') && (d != 'A')); int k = 0; while(k < 4){ c = fgetc(infile); k++; } //RESET OUR INSTRUCTION COUNTER I_COUNTER = 0; STACK_POINTER = 240; } Thanks!

#1. What is MM? How is it defined? #2. fgetc returns an int, not a char.

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it is assumed from other parts of your code that MM should have been declared as either a char pointer and then sized, or should have been declared as fixed storage like char MM[10]

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Originally Posted by rogster001 it is assumed from other parts of your code that MM should have been declared as either a char pointer and then sized, or should have been declared as fixed storage like char MM[10] Yeah, but that's where the error is, so assumptions are all off. @ OP: Indeed, what is the declaration of MM, it's not a double pointer by any chance, is it? It's not defined like: Code: char** MM; or Code: char* MM[#]; Is it? Both cases are double pointers, or arrays of pointers (same thing), the bracket operator will dereference ONE, returning a pointer. It assumes if you're setting a pointer equal to a character, you're doing something wrong. Can you copy/paste the line it is declared on?

Originally Posted by Syndacate Is it? Both cases are double pointers, or arrays of pointers (same thing), the bracket operator will dereference ONE, returning a pointer. It assumes if you're setting a pointer equal to a character, you're doing something wrong. They are not the same thing, but they will yield the same problem in this case.

Originally Posted by Adak io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions. Originally Posted by Salem You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much. Outside of your DOS world, your header file is meaningless.

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