Solve the question

Question

#include <stdio.h>
#include <stdlib.h>
int main()
{
int z; // what would be the value of z ,so that the output will '4'
int *x;
x = (int *)malloc(sizeof(int));
char *c;
c = (char *)malloc(sizeof(char));
*x = z;
c = x;
c++;
printf("%d\n", *c);
return 0;
}
// a) 1234 , b)4132 , c)2431 , d)1243

Answer 1

I will assume a little endian system here, as the given values would not work for a big endian system.

Initially

c = x;

sets `c` to the lowest memory address of the byte sequence of `x`, i.e. it refers to the least significant byte. The increment then advances the pointer to the second byte.

Assuming a standard integer representation with 8 bit, for that byte alone to represent 4, only the third bit may be set (since 2^2 = 4). In the context of the whole integer, that would be the 11nth bit, which contributes as 2^10 = 1024 to its overall value.
The least significant byte can represent anything between 0 and 255, so any answer in the range 1024 to 1279 would be correct. This applies to both 1234 and 1243.

Answer 2

Shadow answer is right
But can u please explain it clearly