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main()
{
int *ptr=(int*)malloc(sizeof(int));
*ptr=4;
printf("%d",(*ptr)+++*ptr++);
}
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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 |
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 |
main(){ int a= 0;int b = 20;char x =1;char y =10; if(a,b,x,y) printf("hello"); }
#include<stdio.h> main() { int i=1,j=2; switch(i) { case 1: printf("GOOD"); break; case j: printf("BAD"); break; } }
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