条件运算符用法

Question

Consider the following statement. What will be the value stored in b?

int a=1;
int b = a+=1 ? a+=1 : 10;

I get the answer as 4. Can anyone explain how that works please.

Answer 1

It has to do with precedence. If you examine the following code (with the rightmost a+=1 changed for simplicity):

#include <iostream>

int main (void) {
    int a=1;
    int b = a+=1 ? 7 : 10;
    std::cout << b << std::endl;
    return 0;
}

you will see that the output is 8 , not 7 or 10 .

That's because the statement:

    int b = a+=1 ? 7 : 10;

is being interpreted as:

    int b = (a += (1 ? 7 : 10));

Now, applying that to your case, we get:

    int b = (a += (1 ? a += 1 : 10));

and, in order of execution:

• the rightmost a += 1 (since 1 is true) sets a to 2 .
• the leftmost a += 2 ( 2 is the result of the previous step) sets a to 4 .
• b = 4 ( 4 is the result of the previous step).

Just keep in mind that you can't necessarily rely on that order of evaluation. Even though there is a sequence point at the ? (so that 1 is evaluated fully before continuing), there is no sequence point between the rightmost a += ... and the leftmost a += ... . And modifying a single variable twice without an intervening sequence point is undefined behaviour, which is why gcc -Wall will give you the very useful message:

warning: operation on ‘a’ may be undefined

That fact that it gives you 4 is pure coincidence. It could just as easily give you 3 , 65535 or even format your hard disk to teach you a lesson :-)

Answer 2

As stated in the other answers these two code snippets are equivalent due to the grammar rules of C++ which determine how compound expressions must be parsed.

int a=1;
int b = a+=1 ? a+=1 : 10;

and

int a=1;
int b = (a += (1 ? (a += 1) : 10));

Although there is a sequence point in a conditional-expression it is between the evaluation of the first expression ( 1 ) and the evaluation of whichever one of the second and third expressions is evaluated ( a += 1 in this case). There is no explicit extra sequence point after the evaluation of the second or third expression.

This means that a is modified twice in the initializer for b without an intervening sequence point so the code has undefined behavior .

Answer 3

Assembly analysis:

int main() 
{
    int a=1;
    int b = a+=1 ? a+=1 : 10;
    return 0;
}

Assembly code generated (using MinGW) for the above code is shown below. The comments are mine, of course! Read the comments also!

 call ___main        //entering into main()
 movl $1, 12(%esp)   //int a = 1; means 12(%esp) represents a;
 incl 12(%esp)       //a+=1 ; a becomes 2
 movl 12(%esp), %eax //loading 'a' onto a register(eax); eax becomes 2 
 addl %eax, %eax     //adding the register to itself; eax becomes 4
 movl %eax, 12(%esp) //updating 'a' with the value of eax; 'a' becomes 4
 movl 12(%esp), %eax //this step could be optimized away; anyway it loads value of 'a' onto the register(eax); eax becomes 4, in fact even earlier it was 4 too! needless step!
 movl %eax, 8(%esp)  //loading the value of eax at another memory location which is 8(%esp); this location represents b; 
 movl $0, %eax       //making eax zero! the return value of main()!
 leave               //now main() says, please leave me!

12(%esp) represent memory location of a , and at a distance 4 byte from it, that is, 8(%esp) represents b . At the end, the value at both these memory locations is 4.

Hence, b = 4. Also a = 4.