为什么变量声明可以很好地作为 for 循环的条件？

Question

What's the return value/type of a declaration like int i = 5 ?

Why doesn't compile this code:

#include <iostream>

void foo(void) {
    std::cout << "Hello";
}

int main()
{
    int i = 0;
    for(foo(); (int i = 5)==5 ; ++i)std::cout << i; 
}

while this does

#include <iostream>

void foo(void) {
    std::cout << "Hello";
}

int main()
{
    int i = 0;
    for(foo(); int i = 5; ++i)std::cout << i; 
}

Answer 1

The for loop requires condition to be either an expression or a declaration:

condition - either

• an expression which is contextually convertible to bool. This expression is evaluated before each iteration, and if it yields false, the loop is exited.
• a declaration of a single variable with a brace-or-equals initializer. the initializer is evaluated before each iteration, and if the value of the declared variable converts to false, the loop is exited.

The 1st code doesn't work because (int i = 5)==5 is not a valid expression at all. (It's not a declaration either.) The operand of operator== is supposed to be an expression too, but int i = 5 is a declaration, not an expression.

The 2nd code works because int i = 5 matches the 2nd valid case for condition ; a declaration of a single variable with a equals initializer. The value of i will be converted to bool for judgement; which is always 5 , then leads to an infinite loop.

Answer 2

This code acutally complies:

for (foo(); int i = 5 == 5; ++i)

It checks if 5 == 5 and set i to this boolean result (which is 1) -> infinite loop

for(foo(); int i = 5; ++i)

This simply checks the value of i after being set to 5, so... when converted to bool it's always true as well -> infinite loop

Answer 3

The syntax of a for loop in C programming language is as follow:

for ( init; condition; increment ) {
   statement(s);
}

The init step is executed first, and only once. You can use init to declare and initialize any loop control variables.

Next, the condition is evaluated. If it is true, the body of the loop is executed. If it is false, the body of the loop does not execute and the flow of control jumps to the next statement just after the 'for' loop.

After the body of the 'for' loop executes, the flow of control jumps back up to the increment statement. This statement allows you to update any loop control variables.

Considering your first program statement:

 for(foo(); (int i = 5)==5 ; ++i)std::cout << i;

The above statement will give you following error (compiled using GCC):

    error: expected primary-expression before ‘int’
        for(foo(); (int i = 5) == 5; ++i)std::cout << i; 

    error: expected ‘)’ before ‘int’

Primary Expression in C can be any of quite a number of things:

a name (of a variable or a function)
a typename
an operator
a keyword like if or while
the list goes on and on ...

This error is due to the fact that statement (int i = 5) == 5 is not a valid expression and compiler is trying to interpret in some other way. However, the second program works as int i = 5 declaration can be used as a valid condition for condition checking in for loop as the such declaration evaluates to value of initializer which is 5 in your second program. However, the second program will be an infinite loop and will keep printing 5 to the terminal.However, if the value of initializer was 0 , then the condition evalautes to false at it's first check and for loop body will not be executed even once.

Answer 4

Although according to a for loop's syntax the initialization statement should come first, and then the checking and then the updation, since your checking statement is an initialization in the second example, i is initialized as 5 as part of the checking and the for loop is continued.

Using the foo() call in initialization makes no difference to the for loop as the return type is void, if you put 0 in place of foo(), you will get the same output

Now let's consider the first example , your checking statement is an initialization, just like the second example, and it would work if you didn't have those brackets there. Since a declaration doesn't return any value, (int i = 5) doesn't return any value and hence cannot be compared with 5 in " == 5 ".

However, if you try "for (foo(); int i = 5 == 5; ++i)..." , the checking statement compiles the same as int i = (5 == 5) . Since 5 == 5 is true, it returns a value of 1, which is then initialized as the value of i. Thus the output is 11111......

TL;DR : Since a declaration doesn't return any value, (int i = 5) doesn't return any value and hence cannot be compared with 5 in " == 5 " as in the first example

Problem solved.