相关疑难解决方法(0)

编译时

void ambig(  signed long) { }
void ambig(unsigned long) { }

int main(void) { ambig(-1); return 0; }

我明白了

error C2668: 'ambig' : ambiguous call to overloaded function
    could be 'void ambig(unsigned long)'
    or 'void ambig(long)'
while trying to match the argument list '(int)'

我知道我可以通过说-1L而不是-1,但为什么/如何在一开始就被认为是模棱两可的来"解决"它？