How to declare a variable in the brackets of if statement?

Question

I want to declare a local variable in the brackets of an if statement. For example.

if((char c = getc(stdin)) == 0x01)//This is not OK with g++.
{
    ungetc(c, stdin);
}

What I want is, to see if the character is the one I want. To say it commonly, I want to use the variable(char c) both in the line of if and the body of if, but not outside the if.

But g++(GCC 4.8.1) says expected primary-expression before 'char'. I wonder if there's a way to do that, because I don't want something like

char c = getc(stdin);
if(c == 0x01)
{
    bla...
}

Answer 1

If it's the namespace pollution you are worrying about you can always define the if statement within a block:

{
    char c = getc(stdin);
    if(c == 0x01)
    {
        // ...
    }
}

So that c will only last until the end of the block is reached.

Answer 2

I didn't know how to create a variable and test its value with an if until after seeing some of the posted solutions. However, you could use switch. This would allow you to react to additional values (perhaps EOF):

switch (int c = getc(stdin)) {
case 0x01: ungetc(c, stdin); break;
case EOF:  // ...handle EOF
default:   break;
}

You could always place the if statement in an inlined function instead, and the code will look a little cleaner. If you really want the source code right at that location, but without creating a new scope around an if with a new variable, then perhaps a lambda would be acceptable to you.

[](int c){ if (c == 0x01) ungetc(c, stdin); }(getc(stdin));

Since you are only comparing against one valuem your particular problem does not require a variable at all, so you can simply do:

if (getc(stdin) == 0x01) {
    char c = 0x01;
    ungetc(c, stdin); //or bla...
}

If you are wanting to compare against a set of values, then the switch suggestion is the better option.

Jerry Coffin's solution looks appealing, but it really boils down to:

if (int c = (getc(stdin) == 0x01)) //...

This is probably not what you really wanted, as it does not generalize well if you want to compare to a value different from 0x01.

Potatoswatter's solution seems closer to what you want, but perhaps it would be nicer to pull the type out into a standalone class:

template <typename T>
class SetAndTest {
    const T test_;
    T set_;
public:
    SetAndTest (T s = T(), T t = T()) : set_(s), test_(t) {}
    operator bool () { return set_ == test_; }
    operator bool () const { return set_ == test_; }
    operator T & () { return set_; }
    operator T () const { return set_; }
};

//...
if (auto c = SetAndTest<int>(getc(stdin), 0x01)) {
    ungetc(c, stdin); //or bla...
}

Answer 3

You can define the variable inside the if statement just fine. For example, this should compile:

if (int ch = getchar())
    ;

The problem is that the type (e.g., int) must follow immediately after the opening parenthesis. The extra parenthesis you have is what's causing compilation to fail. So, if you really want to do this, you'll need to get a little clever and use something like this:

if (char ch = 0 || ((ch = getchar()) == 0x1))

This lets you get the creation and initialization of ch done, then after that part of the expression is complete, put in the parentheses around the ch=getchar() to override the precedence of assignment vs. comparison.

Note that && and || do short-circuit evaluation, so you need to be careful with your initialization. You can use either:

if (char ch = 0 || ...

...or:

if (char ch = 1 && ...

...but if you try to use if (ch = 1 || ... or if (ch = 0 && ..., the short-circuit evaluation will keep the right operand (the part you really care about) from being evaluated at all.

Now the caveat: while I'm reasonably certain this code fits the standard's requirements, and most (all?) current compilers will accept it, it's likely to cause most programmers reading the code some serious head-scratching figuring out what you've done, and why. I'd be extremely hesitant (at best) about using this "technique" in real code.

Edit: It's been pointed out that the result from this may be even more misleading than some initially expect, so I'll try to clarify the situation. What happens is that a value is read from input. That value is assigned to ch and compared to 0x1. So far so good. After that, the result of the comparison (converted to an integer, so either 0 or 1) will be assigned to ch. I believe it has sufficient sequence points that the result is defined behavior. But it's probably not what you, or anybody, want -- thus the advice that you probably don't want to use this, and the mention that it would probably leave most programmers scratching their heads, wondering what you were trying to do. In the very specific case of comparing to 0x1, the value of ch inside the if statement will be 1, but it's more or less a coincidence. If you were comparing to 0x2, the value of ch inside the if would still be 1, not 2.