CODEWITHSUNDEEP
c
ZeroVash
ZeroVash

Reputation: 558

c - initialization makes integer from pointer

I am trying get number ASCII for letter in char variable. My code so far is: 

 int main (int argc,char* argv[]) {
    char c="A"; //here error
    char b="'"; //here error
    char d="z"; //and here error
    printf("%d\n %d\n %d\n",c,b,d); 
}

But I get the following errors:

analizer.c: In function ‘main’:
analizer.c:13:8: warning: initialization makes integer from pointer without a cast [enabled by default]
analizer.c:14:8: warning: initialization makes integer from pointer without a cast [enabled by default]
analizer.c:15:8: warning: initialization makes integer from pointer without a cast [enabled by default]

Upvotes: 2

Views: 1958

Answers (3)

Anish Sharma
Anish Sharma

Reputation: 505

In C strings are just an array of characters which terminate with a \0. So if you want to use only one character, declare a char variable and use the character single quote. If you want to use a string, declare char array and use doublequote as

char a='A' //  just one character
char b[20]="Some_Text"// multiple characters(a string)

In the first case, a contains the integer value of 'A' but in the second, b contains the base address of the string it points to. Each character in array b[] must be accessed with indexes as b[0] b[1] etc.

You can use strings to represent single characters as

char a[]="A" // a string of size 1

and then you can print it's integer equivalent as

printf("%d",a[0]); //0th element of the string

or

use the single quote method as described in other answers.

Upvotes: 1

MikeCAT
MikeCAT

Reputation: 75062

"A" is a string literal (an array of characters). 'A' is a character constant (an integer). What you want should be latter.

#include <stdio.h>

int main (int argc,char* argv[]) {
    char c='A';
    char b='\''; // use escape sequence
    char d='z';
    printf("%d\n %d\n %d\n",c,b,d); 
}

The compiler will allow you to extract elements of the arrays (string literals) like below, but using character constants should be better in this case.

#include <stdio.h>

int main (int argc,char* argv[]) {
    char c="A"[0]; // a typical way to deal with arrays
    char b=*"'"; // dereferencing of a pointer converted from an array
    char d=0["z"]; // a tricky way: this is equivalent to *((0) + ("z"))
    printf("%d\n %d\n %d\n",c,b,d); 
}

Upvotes: 6

Linus
Linus

Reputation: 1518

Basically you point to a string literal when you do this (which is invalid):

char c="A";

You need to write a single char to the variables:

char c='A';
char b='\'';
char b='z';

Upvotes: 2

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