Why existing function arguments cannot be used to evaluate other default arguments?

Question

I was writing a function foo() which takes 2 const char*s as arguments, pBegin and pEnd. foo() is passed a null terminated string. By default pEnd points to \0 (last character) of the string.

void foo (const char *pBegin,
          const char *pEnd = strchr(pBegin, 0))  // <--- Error
{
  ...
}

However, I get an error at above line as:

error: local variable ‘pBegin’ may not appear in this context

Why compiler doesn't allow such operation ? What's the potential problem ?

Answer 1

The standard not only explicitly disallows the use of other parameters in a default argument expression, but also explains why and gives an example:

ISO/IEC 14882:2003(E) - 8.3.6 Default arguments [dcl.fct.default]

9. Default arguments are evaluated each time the function is called. The order of evaluation of function arguments is unspecified. Consequently, parameters of a function shall not be used in default argument expressions, even if they are not evaluated. Parameters of a function declared before a default argument expression are in scope and can hide namespace and class member names. [Example:

    int a;
    int f(int a, int b = a);         // error: parameter a
                                     // used as default argument
    typedef int I;
    int g(float I, int b = I(2));    // error: parameter I found
    int h(int a, int b = sizeof(a)); // error, parameter a used
                                     // in default argument

—end example] ...