Answer / jaya.vavilala

output is error.because we cannot inrease address of
array.if we increase address can be increased after go on
increasing after 5th position it cannot find where it is
present.so we cannot increase array directly through
pointer we can increase.

Answer / vikas thakur

The answer is error.
The reason is that we can't increment a constant(the
variable int a[] ). The expression int a[] is the address
where the system had placed your array and it will remain
to stay at that address until the program terminates. you
can't increment an address but you can increment a pointer.

.....the correct program would be....

void main()(
int a[]={1,2,3,4,5},i;
for(i=0;i<5;i++)
printf("%d",a[i]);
getch();
}

....in another way.....

void main()(
int a[]={1,2,3,4,5},i;
int *p;
p = a;
for(i=0;i<5;i++)
printf("%d",*(p++));
getch();
}

Answer / chhaya

Here , in program array is given of 5 element. But at d
time of printing output ,increase address not pointer
i.e.
a[i]={1,2,3,4,5}
then u wnat to print array
Printf("%d",i++);
getch();

this sentence is requird. because array is set of element
and element is pointed by "pointer". If u wnt to show
element then use pointer not address of array.....

1) int main() { unsigned char a = 0; do { printf("%d=%c\n",a,a); a++; }while(a!=0); return 0; } can anyone please explain the explain the output

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