Thread: to pass address from char to pointer VS to pass address from integer to pointer

Hi guys,

Whenever I pass an adress from interger variable to a pointer (a function argument) I need to put '&' command, but when I pass an address from char variable to a pointer(function argument too) I don't need to put '&'. Why does it happen?

Code:

char text[3] = "hi ";
int v = 0;

void func1(char *str);
void func2(int *x);

void main()
{
   func1(text);
   func2(&v);
}

void func1(char *str)
{
while(*str==' ')
{
*str='!';
}
}

void func2(int *x)
{
*x = 1;
}

You're not passing a single character to the function, you're passing a character array. Arrays are passed to functions as pointers, and an array name by itself acts as a pointer to the first element in this context. Hence, since you're already passing a pointer, you don't need the &.

Question 6.4

Note that this does not mean pointers and arrays are the same - they just behave similarly sometimes:

Question 6.3

Also:

• You should declare your variables in "main()" instead of globally
• "main()" returns int, not void: FAQ > main() / void main() / int main() / int main(void) / int main(int argc, char *argv[]) - Cprogramming.com
• In the function "func1()", you're not incrementing the pointer.