1)#include <iostream.h>
int main()
{
int *a, *savea, i;
savea = a = (int *) malloc(4 * sizeof(int));
for (i=0; i<4; i++) *a++ = 10 * i;
for (i=0; i<4; i++) {
printf("%d\n", *savea);
savea += sizeof(int);
}
return 0;
}
2)#include <iostream.h>
int main()
{
int *a, *savea, i;
savea = a = (int *) malloc(4 * sizeof(int));
for (i=0; i<4; i++) *a++ = 10 * i;
for (i=0; i<4; i++) {
printf("%d\n", *savea);
savea ++;
}
return 0;
}
The output of this two programs will be different why?
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Answer Posted / ips
In 1st Case(savea++)
--------------------
The Integer Pointer Is Incremented just Once.(as it Is
Implimented in c/c++).which means the pointer is shifted
4bytes(size of type 'int') ahead.
In 2nd Case(savea+=sizeof(int))
-------------------------------
Here The Statement implies:-
savea+=4;
The above statement says that,the integer pointer is to
be increamented 4times.means,the Pointer now is shifted 16
bytes(4*sizeof type 'int').which is Out of scope of the
integer array in the Programme.
| Is This Answer Correct ? | 3 Yes | 1 No |
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