C error: Expected expression before int

This is actually a fairly interesting question. It’s not as simple as it looks at first. For reference, I’m going to be basing this off of the latest C11 language grammar defined in N1570

I guess the counter-intuitive part of the question is: if this is correct C:

if (a == 1) {
  int b = 10;
}

then why is this not also correct C?

if (a == 1)
  int b = 10;

I mean, a one-line conditional if statement should be fine either with or without braces, right?

The answer lies in the grammar of the if statement, as defined by the C standard. The relevant parts of the grammar I’ve quoted below. Succinctly: the int b = 10 line is a declaration, not a statement, and the grammar for the if statement requires a statement after the conditional that it’s testing. But if you enclose the declaration in braces, it becomes a statement and everything’s well.

And just for the sake of answering the question completely — this has nothing to do with scope. The b variable that exists inside that scope will be inaccessible from outside of it, but the program is still syntactically correct. Strictly speaking, the compiler shouldn’t throw an error on it. Of course, you should be building with -Wall -Werror anyways 😉

(6.7) declaration:
            declaration-speciïŹers init-declarator-listopt ;
            static_assert-declaration

(6.7) init-declarator-list:
            init-declarator
            init-declarator-list , init-declarator

(6.7) init-declarator:
            declarator
            declarator = initializer

(6.8) statement:
            labeled-statement
            compound-statement
            expression-statement
            selection-statement
            iteration-statement
            jump-statement

(6.8.2) compound-statement:
            { block-item-listopt }

(6.8.4) selection-statement:
            if ( expression ) statement
            if ( expression ) statement else statement
            switch ( expression ) statement

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