Why the same variable print the different output?

Question

#include <stdlib.h>
#include <stdio.h>
void roll_three(int* one, int* two, int* three)
{

  int x,y,z;
  x = rand()%6+1;
  y = rand()%6+1;
  z = rand()%6+1;

  one = &x;
  two = &y;
  three = &z;
  printf("%d %d %d\n", *one,*two,*three);  
}
int main()
{
  int seed;
  printf("Enter Seed: ");
  scanf("%d", &seed);
  srand(seed);
  int x,y,z;
  roll_three(&x,&y,&z);
  printf("pai: %d %d %d\n", x,y,z);
  if((x==y)&&(y==z))
    printf("%d %d %d Triple!\n",x,y,z);
  else if((x==y)||(y==z)||(x==z))
    printf("%d %d %d Double!\n",x,y,z);
  else
    printf("%d %d %d\n",x,y,z);
  return 0;

}

This the terminal, I type 123 for the seed. However, the printf in roll_three and the printf in main give me different output? Why *one and x are different?

the function srand() is expecting an unsigned int so suggest main() start with: unsigned seed; printf("Enter Seed: "); scanf("%u", &seed); srand(seed); — user3629249
– user3629249, Commented Feb 27, 2016 at 21:06

Schwern · Accepted Answer · 2016-02-26 22:47:39Z

6

The problem is here:

one = &x;
two = &y;
three = &z;

Since one, two, and three are pointers you've changed what they point to, but now they no longer point to main's x, y and z. It is similar to this...

void foo(int input) {
    input = 6;
}

int num = 10;
foo(num);
printf("%d\n", num);  // it will be 10, not 6.

Instead you want to change the value stored in the memory they point to. So dereference them and assign them the new value.

*one = x;
*two = y;
*three = z;

You can even eliminate the intermediate values.

*one   = rand()%6+1;
*two   = rand()%6+1;
*three = rand()%6+1;

answered Feb 26, 2016 at 22:47

Schwern

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Comments

Tom Karzes · Accepted Answer · 2016-02-26 22:37:03Z

2

In roll_three, you need to change the following:

one = &x;
two = &y;
three = &z;

To:

*one = x;
*two = y;
*three = z;

The first version just points to them to the local variables. The corrected version updates the values in the caller.

answered Feb 26, 2016 at 22:37

Tom Karzes

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4 Comments

Hanslen Chen Over a year ago

But when I make the int x,y,z in roll_three as a global variable, write behind the #include<stdio.h>, there are still different? I know "*one = x;" is correct, but why "one = &x;" is not correct.

Tom Karzes Over a year ago

If you make them global, then you have to remove the local declarations from both roll_three and main. Otherwise the local definitions will mask the global ones. But you don't really want to make them global.

Hanslen Chen Over a year ago

I rename that x, y,z as xx, yy, zz and change the all x, y,z in roll_three as xx yy zz. There still occur some problem, I will post that in another comment, can you check that for me?

Tom Karzes Over a year ago

I'm not sure what you're doing at this point, but if you post the code I can take a look.

tdao · Accepted Answer · 2016-02-26 22:39:07Z

1

Why *one and x are different?

one = &x;

should be

*one = x;

The way you did it ( one = &x ) is wrong, because you assign pointer one to the address of local variable x which no longer exists after function roll_three.

You function should look like this:

void roll_three(int* one, int* two, int* three)
{

    int x,y,z;
    x = rand()%6+1;
    y = rand()%6+1;
    z = rand()%6+1;

    *one = x;
    *two = y;
    *three = z;
    printf("%d %d %d\n", *one,*two,*three);  
}

edited Feb 26, 2016 at 22:39

answered Feb 26, 2016 at 22:36

tdao

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2 Comments

Hanslen Chen Over a year ago

But when I make the int x,y,z in roll_three as a global variable, write behind the #include<stdio.h>, there are still different? I know "*one = x;" is correct, but why "one = &x;" is not correct.

tdao Over a year ago

@HanslenChen - it can't be; unless you declared x,y,z as global variables, and then forgot to delete the local variables x,y,z in main

Kaizhe Huang · Accepted Answer · 2016-02-26 22:45:12Z

0

In your roll_three function,

void roll_three(int* one, int* two, int* three)
{
  int x,y,z;
  x = rand()%6+1; // local variable x will have a value
  y = rand()%6+1;
  z = rand()%6+1;

  one = &x; // variable one is assigned with local variable x's address
            // so *one equals to x inside the function.
            // However, variable one supposed is the variable x's address in 
            // the main function, but now it is changed to the local x's address,
            // the main function variable x's value can't be updated as expected.
            // *one = x is the way you want to go.
  two = &y;
  three = &z;
  printf("%d %d %d\n", *one,*two,*three);  
}

answered Feb 26, 2016 at 22:45

Kaizhe Huang

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