Explain the output of the following C Code.

Question

CODE :
#include <stdio.h>
int main()
{
char arr[10] = {'a','b','\0','d'};
printf("%s\n",arr);

if(arr == "ab") printf("true");
else printf("false");
}

OUTPUT :
ab
false

[Why not :
ab
true]

Answer 1

Because arr == "ab" compares the pointer to the second ab, which is stored in the .data segment, and the pointer to arr.

The strcmp function from string.h compares 2 strings. It returns 0 if they are equal.

if (strcmp(arr, "ab") == 0) should give you the desired result.

Answer 2

Two reasons:

A) when you say arr == x, you are comparing the address of the first value in arr (not the contents of arr) to x.

B) even if == was a .compareTo function, it would look loop through every char in arr, which includes 'd'. The print string abikity in printf just prints until the first null character.