CODEWITHSUNDEEP
c
Blue-Maned Hawk
Blue-Maned Hawk

Reputation: 175

Recieving warning about casting "from pointer to integer of different size" while compiling c in gcc

I am trying to learn C, and I'm trying to cast a char into an int to extract the ASCII value. However, when I try to compile this in GCC, I get this warning:

warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]
     ascii = (int) letter;

I'm trying to do a simple program that asks for a single character, scans for it, then casts it as an int to get the ASCII value, printing that value. I've initialized the letter variable as char* and ascii as int. I've tried using the placeholders%s, %c, and %1s for the letter variable, but it isn't work.

Here is the code:

#include <stdio.h>

char* letter;
int ascii;

int main(){
    printf("Please input a charcter:");
    scanf("%s", letter);
    ascii = (int) letter;
    printf("\n The ASCII value of %s is %d.\n", letter, ascii);
}

What I'd expect to happen is that "Please input a character:" would print, then a character, such as a, would be typed, and then it would print, for example, `"The ASCII value of a is 97."

Instead, when I type something, it prints "The ASCII value of (null) is zero." During compilation, it prints out the error listed above. This is not what's supposed to happen.

How can I fix this problem?

Upvotes: 0

Views: 264

Answers (3)

user17791426
user17791426

Reputation:

Just use %hhd format specifier

#include <stdio.h>
int main(void)
{
    char letter;
    fputs(stdout, "Please input a charcter:");
    letter = getchar();
    
    printf("\nThe ASCII value of %c is %hhd.\n", letter, letter);
    return 0;
}

Upvotes: 0

AaronHolland
AaronHolland

Reputation: 1685

You aren't casting a char to an int, you are casting a pointer to a char to an int.

int main(){
    // define as an actual char, not a char*
    char letter;
    int ascii;

    printf("Please input a character:");
    // scan a character, not a string.  Pass in the address of the char
    scanf("%c", &letter);
    ascii = (int)letter;
    printf("\n The ASCII value of %c is %d.\n", letter, ascii);
}

Upvotes: 1

Bwebb
Bwebb

Reputation: 685

#include <stdio.h>

//char* letter; //moving to stack from data segment
int ascii;

int main(){
    char letter[10] = {0};//allocing 10 bytes here, all initialized null chars
    printf("Please input a charcter:");
    scanf("%c", &letter); //not checking return value here, bad form
    ascii = (int) letter[0]; //casting only the first char
    printf("\n The ASCII value of %s is %d.\n", &letter, ascii);
}

Results in the output that you wanted:

Please input a charcter:
 The ASCII value of a is 97.

This is bad code, and ive commented on some lines to help add some clarity. 

#include <stdio.h>

//char* letter; //moving to stack from data segment
int ascii;

int main(){
    char letter[10] = {0};//allocing 10 bytes here, all initialized null chars
    printf("Please input a charcter:");
    scanf("%s", &letter); //not checking return value here, bad form
    ascii = (int) letter[0]; //casting only the first char
    printf("\n The ASCII value of %s is %d.\n", &letter, ascii);
    printf("\n The ASCII value of %s is %d.\n", &letter[1], (int)letter[1]);
}

With input ab, changing %c to %s gets both charaters (which seems like what you wanted in the first place?)

Please input a charcter:
 The ASCII value of ab is 97.


 The ASCII value of b is 98

.

Upvotes: 0

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