I was learning about pointers and pointers to arrays, when I encountered this strange thing.
Can anyone explain why this works?
char str[] = "This is a String";
int *p = str;
while(*p) std::cout<<*p++;
and returns :
but this one generates a nasty error:
int arr[] = {1,2,3,4,5,6,7,8,9};
int *p = arr;
while(*p) std::cout<<*p++<<" ";
like this :
I know that this is undefined behavior, so is there any way to resolve this?
char str[] = "This is a String"; int *p = str;
This program is ill-formed. char[N] does not implicitly convert to int* in C++, only to char*.
int arr[] = {1,2,3,4,5,6,7,8,9}; int *p = arr; while(*p) std::cout<<*p++<<" ";I know that this is undefined behavior
Do you understand why it is undefined behaviour? The problem is that the condition to end your loop is when you reach the element with value 0. But you didn't put any zeroes in your array, so the loop never stops and you access beyond the end of the array.
is there any way to resolve this.
Sure. Just don't access beyond the bounds of the array.
A good choice is to use a range-for loop:
for(int i : arr) {
std::cout << i << ' ';
}
whileloop won't stop until it sees a 0. In the first example, all string literals have a terminatingnulcharacter which is 0.whileloop to compare by size, not by content.char[]to anint*. But even if it did compile, it would be printingintvalues, notcharvalues. So the only way you could be gettingThis is a Stringas output is if you were using achar*instead of anint*