如果在函数中使用，“return b ++”中“b”的增量会发生什么变化？

Question

Since b++ is post-increment,what happens to the increment of b if used as return b++ as in the following program?

#include<stdio.h>

int foo(int);

int main()
{
   int a=8;
   printf("%d",foo(a));
}

int foo(int a)
{
   static int  b=a*a;
    return b++;
}

Edit

#include<stdio.h>
int foo();

int main()
{
foo();
foo();
}

int foo()
{
    static int b=1;
    printf("%d\n",b);
    return b++;
}

Result

1

2

As I saw in my edit, why is b incremented at all?Isn't return supposed to exit that function immediately?Why is b incremented even after control returns to main() ? Aren't all activities in the function supposed to end after return?

Answer 1

Many C (sub-)expression have a value and a side effect.

The value of b++ is the value of b before evaluating the expression; its side effect is to increase the value in b by one.

So, the expression return b++; returns the previous value of b and updates b . When b is static the update stays around for the next function call; when b is a plain old local variable, the update is lost (a smart compiler will even not emit code to update the object).

Answer 2

Actually, this code doesn't compile, because b 's initializer isn't a constant.

Since b is static, it's essentially a global variable, and it gets initialized before main , at which point a doesn't exist.

If on the other hand we compile this as C++, b is initialized the first time that foo is called - so gets the value 64. At the return, b++ is incremented and stored as 65 - but the return value is 64. So if you call foo again, it will return 65 (and b is 66).

Edit based on edited code:

So, the code essentially performs:

 int temp = b;
 b = b + 1;
 return temp; 

This is the way that C is define. The result of x++ is the previous value x , but the value of x is increment. Since static is essentially a global (but without a name, so you can't use it outside of that function), the value persists between calls to the function, and it is initialized before main is called.

Answer 3

return b++; is equivalent to: int c = b; b = c + 1; return c; int c = b; b = c + 1; return c;

To answer your new questions: - The function exits after return indeed, but before it returns, the expression b++ must be evaluated first. Evaluating such an expression will result in b being incremented. - All activities in the function "end" after return but b is declared as a static variable, in which case its value persists through subsequent executions of the function.

To make it easier for you to understand, a post increment is like the following function

int postincrement(int& x) {
    int y = x;
    x = x + 1;
    return y;
}

(Of course the compiler could optimize a b++ if it finds that incrementing b has no effect at all, but in this case it does have an effect (increment the static int b ) so it can't be optimized out.)

Answer 4

This post-increment is completely useless - because b has local scope, its incremented value is disposed after return . If your compiler is smart, it most likely to optimize it out.