Question

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Question about pointers and strings in C

#include<stdio.h>

int main()   
{    
char *str1="abcd";  
char str2[]="abcd";  
printf("%d %d %d\n",sizeof(str1),sizeof(str2),sizeof("abcd"));  
return 0;
}

Why does this code give same answers for sizeof(str2) and sizeof("abcd") even when str2 is ideally just like a pointer to a string , as is str1 ,so answer should be 4 4 5

Code on Ideone:
http://ideone.com/za8aV

Answer: 4 5 5

Answer 1

Where did you get the idea that str2 is “ideally just like a pointer to a string”? It is not. str2 is an array. When operator sizeof is applied to an array, it returns the size of the array object in bytes.

String literal is also an array, so when sizeof is applied to a string literal, it returns the size of that array object in bytes. So, it is perfectly natural to expect sizeof("abcd") and sizeof(str2) to produce the same result. And they do.

P.S. %d is not an appropriate format specifier to print the result of sizeof. %d requires int argument, while sizeof produces a size_t value. Use %zu to print values of size_t type.