CODEWITHSUNDEEP
cendianness
rodrigues
rodrigues

Reputation: 13

Endianness issue?

What will be the output of the following code?

#include<stdio.h>
int main()
{
    int a[4]={4096,2,3,4};
    char *p;
    p=a;
    printf("\n %d",(int)*(++p));
    return 0;
}

sizeof int = sizeof(void*) = 4 bytes

According to me the output should be 16 on a little endian machine and 0 on a big endian machine. Am I right?

Upvotes: 1

Views: 467

Answers (2)

Steve Jessop
Steve Jessop

Reputation: 279405

4096 is 0x1000, so (once you get it to compile) you'd expect 16 with a little-endian representation, and 0 with a big-endian representation unless int is 24 bits wide (that would give 16).

All this assuming CHAR_BIT == 8, and no funny-business with padding bits. The edit with sizeof(int) == 4 also rules out the 24 bit int under those assumptions.

Upvotes: 1

unwind
unwind

Reputation: 400129

4096 (hex 0x1000) will be represented in memory (assuming 8-bit bytes, which is quite common nowadays) as either:

[0x00, 0x00, 0x10, 0x00] (big-endian)

or

[0x00, 0x10, 0x00, 0x00] (little-endian)

Since you're printing it out using %d, which expects integers, but passing a dereferenced character pointer value (i.e., a char) and incrementing the pointer before dereferencing it, you will print either 0x00 or 0x10, which is 16 in decimal. You will probably need to add some cast to allow the p = a statement, since you're mixing types rather freely.

So yes, I think you're right.

Upvotes: 1

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