c - 将超过要求的参数传递给 c 函数

Question

如果我向函数传递的参数多于所需参数，会发生什么情况？我预计被调用函数中的某些内容会被破坏，但在一些小的测试代码中一切正常。

例如:

void print()
{
    int x=10;
    printf("%d\n",x);
}
void main()
{
    print(0,0,0,0,0);
}

Answer 1

这是未定义的行为。

(C99, 6.5.2.2p6) "If the expression that denotes the called function has a type that does not include a prototype, [...] If the number of arguments does not equal the number of parameters, the behavior is undefined."

并且我们从 6.9.1p7 中得知 print 函数不提供原型(prototype)。

C99, 6.9.1p7) "If the declarator includes a parameter type list, the list also specifies the types of all the parameters; such a declarator also serves as a function prototype for later calls to the same function in the same translation unit. If the declarator includes an identifier list,142) the types of the parameters shall be declared in a following declaration list."

由于没有违反约束，因此不需要诊断消息。