Hello I have this code that fills the array with a pointer and print it with a pointer as well. The result is faulse. Here is my code:
#include<stdio.h>
#include<stdlib.h>
int main(){
int p;
char po[8];
char *i;
i=po;
for(p=0; p<8; p++){
scanf("%c\n", i++);
}
for(p=0; p<8; p++){
printf("%c\n", *(i++));
}
return 0;
}
where is my fault?
There can be good reasons for increasing pointers. Simply do not forget to reset them to their initial value before reusing them!
Your code should be:
i=po;
for(p=0; p<8; p++){
scanf("%c\n", i++);
} /* i is now po + 8 ... */
i = po;
for(p=0; p<8; p++){
printf("%c\n", *(i++));
}
Now you should learn not to write such code:
i is already pointing at the end of the array. You want to print po using p as index:
for(p=0; p<8; p++){
printf("%c\n", po[p]);
Also, you don't need the \n in the scanf() call. Any whitespace character in format specifier will ignore all whitepaces in the input and as such you will need to input a non-whitespace character at the end to end the input.
The problem is that the pointer i being incremented in the first loop, will point to the end of the array at the end of the loop (more precise, at the next memory location after the one allocated for po, i.e. will point to po[8] which is not part of the array). You can add
i = po;
before the second loop, to make it point again to the beginning of po (to po[0]).
scanf("%c\n", i++);...please don't write this sort of code.