将字符串文字分配给char变量时，为什么打印字符d？

Question

#include<stdio.h>

int main(void) {
    char a = "any"; //any string
    printf("%c", a);
    getch();
}

Why always d (for %c ) or 100 (for %d ) gets printed? What's happening?

Answer 1

char a="any"; is not declaring any string. Your compiler should through error/warning. You need

const char *a = "any";   

Now

printf("%c", *a);

will print character a as pointer a is pointer to the first element of the string literal any .

Answer 2

Your code has issues. On

char a="any";

the string literal "any" is a pointer but you are saving it in a (small) integer type. Because the value of a is (part of) an address it will have an essentially arbitrary value. On my machine it prints "T" (if I remove the getch() line because that doesn't compile).

GCC gives the following warning:

warning: initialization makes integer from pointer without a cast

and whatever compiler you are using probably tells you something similar. You should really take this warning seriously.

Answer 3

"something between double quotes" 

Allocates memory for the string and returns a pointer to the string in memory. you need to store this pointer in a using char *a = "abc"; now 'a' is a pointer to the string abc.

printf("%s",a) this will print the entire string abc.

for a single character use single quotes 'a' in this case char a = 'b'; is correct.

printf("%c",a) this will print b.
printf("%d",a) this will print the ascii value of 'b', which is 98

Answer 4

In the statement

char a = "any";

the string literal "any" evaluates to the address of its first character which is of type const char * . Assigning a const char * to a char variable implicitly converts the pointer to an integer. This emits the following warning if you compile with -W gcc flag:

warning: initialization makes integer from pointer without a cast [enabled by default]

What happens next is the least significant byte of the cast integer value is assigned to the char variable a . This is a random value and cannot be predicted. On my machine, it happened to be the ascii code for the character / . On your machine, it happened to be 100, the ascii code for the character d . It is just pure chance. Nothing else.

You should assign a string literal to a const char * as:

const char *a = "any";

and print the string pointed to by a by the %s conversion specifier as:

printf("%s\n", a); 

Also note that getch is not a C standard library function, so your program won't compile on a linux machine. You can use getchar for the same effect, though.