difference between *y++ and ++*y?

Question

I'm confused in how this code will get executed. Suppose we have ``` int x=30,*y,*z; y=&x; ``` what is the difference between *y++ and ++*y? and also what will be the output of this program? ``` #include<stdio.h> int main(){ int x=30,*y,*z; y=&x; z=y; *y++=*z++; x++; printf("%d %d %d ",x,y,z); return 0; } ```

Accepted Answer

The expression `x = *y++` is in effects same as:

```
x = *y;
y = y + 1;
```

And if expression is just `*y++;` (without assignment) then its nothing but same as `y++;`, that is `y` start pointing to next location after increment.

Second expression `++*y` means to increment the value pointed by `y` that same as: `*y = *y + 1;` (pointer not incremented) It will be better clear with answer to your first question:

Suppose your code is:

```
int x = 30, *y;
int temp;
y = &x;

temp = *y++; //this is same as:  temp = *y;  y = y + 1;
```

First `*y` will be assigned to `temp` variable; hence `temp` assigned `30`, then value of `y` increments by one and it start point to next location after location of `x` (where really no variable is present).

Next case: Suppose your code is:

```
int x = 30, *y;
int temp;
y = &x;

temp = ++*y;  //this is same as *y = *y + 1; temp = *y;
```

First value of `*y` increments from `30` to `31` and then `31` is assigned to `temp` (note: `x` is now `31`).

next part of your question (read comments):

```
int x = 30, *y, *z;

y = &x;       // y ---> x , y points to x
z = y;        // z ---> x , z points to x
*y++ = *z++;  // *y = *z, y++, z++ , that is 
              // x = x, y++, z++
x++;          // increment x to 31
```

difference between y++ and ++y?

Solution

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