Why Does a C Function Compile Without an Explicit Return Statement Despite Declaring a Return Type?

Why Does This C Snippet Compile Despite Missing a Return Value?

In the code snippet provided:

static tvec4 Min(const tvec4& a, const tvec4& b, tvec4& out)
{
    tvec3::Min(a, b, out);
    out.w = min(a.w, b.w);
}

The function Min is declared as returning a tvec4, but the code does not provide an explicit return statement. This raises the question of why the compiler compiles the code without an error.

Undefined Behavior

According to the C 11 draft standard, flowing off the end of a value-returning function without providing a return value results in undefined behavior. This means that the compiler is not obligated to issue an error or warning in such cases.

Compiler Options

However, compilers can be configured to warn or even generate errors for undefined behavior. For instance, using the -Wall flag can often trigger a warning like:

warning: control reaches end of non-void function [-Wreturn-type]

By adding the -Werror=return-type flag, the compiler will treat this warning as an error, forcing the code to be corrected.

Visual Studio

In Visual Studio, the code will generate error C4716 by default:

error C4716: 'Min' : must return a value

In cases where not all code paths return a value, Visual Studio will issue a warning (C4715).

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