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main()
{
int *ptr=(int*)malloc(sizeof(int));
*ptr=4;
printf("%d",(*ptr)+++*ptr++);
}

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main() { int *ptr=(int*)malloc(sizeof(int)); *ptr=4; printf("%d",(*ptr)+++*ptr++); }..

Answer / john lee

main()
{
int *ptr=(int*)malloc(sizeof(int));
*ptr=4;
printf("%d",(*ptr)++ + (*ptr)++);
}

Like above, code should be revised as allocated memory space just has 2(16bit machine) or 4 byte(32bit machine) to save '4'.
If not, the orginal code, printf("%d",(*ptr)++ + *ptr++);

In my guess, ptr++ will be first, and then *ptr would be next.
If so, ptr++ will point to a memory address unintended(an address +2 or +4 added). And then *ptr will have a value like 0 or else.

Is This Answer Correct ?    0 Yes 0 No

main() { int *ptr=(int*)malloc(sizeof(int)); *ptr=4; printf("%d",(*ptr)+++*ptr++); }..

Answer / ram

error due to constant can't be increment

Is This Answer Correct ?    0 Yes 0 No

main() { int *ptr=(int*)malloc(sizeof(int)); *ptr=4; printf("%d",(*ptr)+++*ptr++); }..

Answer / vinod

Answer = 8
Explanation:
*ptr=4;
printf("%d",(*ptr)+++*ptr++);
The above statement can be interpreted as (*ptr)++ + *ptr++
(*ptr)++ - Post increment the Value pointer by ptr i.3. 4
*ptr++ - Return the value of ptr and increment the position of ptr i.e. 4
So (*ptr)++ + *ptr++ => 4 + 4 => 8

Is This Answer Correct ?    1 Yes 2 No

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