In C, char array and string array in c shows different size, but when compared they are same

Question

In C while trying char arrays, I came up with this problem.

void main(){

    char buffer[5] = {'s','d','f','d','f'};
    char a[5] = "sdfdf";
    printf("%d\n", *a==*buffer);
    printf("%s\n", buffer);
    printf("%d\n", (int)strlen(buffer));
    printf("%d\n", (int)strlen(a));
}

Output is

1

sdfdf

sdfdf

5

6

Agreed, it's pure luck that strlen(buffer) did not tell you 42. — Weather Vane
– Weather Vane, Commented Mar 16, 2015 at 19:46
If some tutorial, book, or instructor advised you to use void main(), find get a better one. Use int main(void). (And you omitted #include <stdio.h>.) — Keith Thompson
– Keith Thompson, Commented Mar 16, 2015 at 20:41

Iharob Al Asimi · Accepted Answer · 2015-03-16 19:41:46Z

4

The main problem is the missing '\0', to define a 5 characters string correctly and initialize it from an array of char with an initializer list this is how it should be

char buffer[6] = {'s', 'd', 'f', 'd', 'f', '\0'};

the second variant in your code si wrong because the array can only store 5 characters, so again

char a[6] = "sdfdf";
/*     ^ 6 instead of 5 */

you cannot expect any of the functions following that part of the code to work, when you have a missing '\0'.

All of the printf() with the "%s" spcifier and strlen() expect this last special value '\0' or 0 if you prefer, to be there, when it's not, then these functions invoke undefined behavior because they go beyond the end of the array searching for the '\0'.

answered Mar 16, 2015 at 19:41

Iharob Al Asimi

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