with the below code what exactly below?
Is the address of 'a' being copied into the address of 'i'? If so, when I state temp=*i; does that mean that the address of a is copied into temp. *i means contents of i which is '&a'
Code:#include <stdio.h> int fun1(int *i,int *j) { int temp; temp=*i; *i=*j; *j=temp; return(0); } int main() { int a,b; scanf_s("%d%d",&a,&b); printf("Before call %d %d \n",a,b); fun1(&a,&b); printf("After first call %d %d \n",a,b); getchar(); getchar(); return(0); }
Last edited by bos1234; 02-14-2011 at 06:36 AM.
No. Think about the type of i.Originally Posted by bos1234
Print these addresses as a test.
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*i is a pointer so technically it's taking the address of a so basically yes the address of a is in i. this means that *i will point to the variable located in the address assigned to it which is a in this case
if you swap i and j's address they won't affect a and b. this will only make the pointers point to different variables in memory. but i see you are actually swapping the values in fun1 by using *i = *j; to swap memory addresses use this
Code:int fun1(int *i,int *j) { int * temp; temp=i; i=j; j=temp; return(0); }
We shouldn't be quick to criticize unless we can do better.
Code:public function __clone() { die ( "I'm one of a kind" ); }
Code:int fun1(int *i,int *j) <--- i = address of a, j = address of b { int temp; <--- integer value temp=*i; <--- temp = value at address of i *i=*j; <--- value at i = value at j *j=temp; <--- value at j = temp return(0); <--- return 0 }
