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我来解释一下那些么明奇妙的WA

Posted by gyarenas at 2011-11-21 15:20:38 on Problem 3295
WA来自那些递归下降求解的代码.
第一种情况,使用|| 和 &&:
例如s为所给串
int getval()
{
	switch(s[c_s++])
	{
	case 'p': return (value & (1 << 0))? 1:0;
	case 'q': return (value & (1 << 1))? 1:0;
	case 'r': return (value & (1 << 2))? 1:0;
	case 's': return (value & (1 << 3))? 1:0;
	case 't': return (value & (1 << 4))? 1:0;

	case 'K': return getval() && getval();
	case 'A': return getval() || getval();
	case 'N': return !getval();
	case 'C': return !getval() || getval();
	case 'E': return getval() == getval();
	}
}
这种情况简单,大家知道短路求值吧,先对 ||左边的表达式求值,如果非0,则不会对右边的表达式求值,&&同理,如果左边为0,不会对右边求值,这样下标就不会按照事先设想的增加
第二种情况,使用&和|:
int getval()
{
	switch(s[c_s++])
	{
	case 'p': return (value & (1 << 0))? 1:0;
	case 'q': return (value & (1 << 1))? 1:0;
	case 'r': return (value & (1 << 2))? 1:0;
	case 's': return (value & (1 << 3))? 1:0;
	case 't': return (value & (1 << 4))? 1:0;

	case 'K': return getval() & getval();
	case 'A': return getval() | getval();
	case 'N': return !getval();
	case 'C': return !getval() | getval();
	case 'E': return getval() == getval();
	}
}
首先这段代码用G++会AC,C++会WA,说明此段代码依赖编译器,是未定义的代码
错误原因: C语言对表达式的求值顺序不是明确规定的,而G++是从左从左向右求值,C++则正好相反,比如: getval() | getval() G++先对左边的getval()求值,而C++则先对右边的getval()求值,也就会导致对s的访问顺序不会按预先的步骤进行,所以用G++会ac,C++会WA掉
正确的写法,去掉那些跟依赖求值顺序的代码(好习惯),分别求值,用两个临时变量保存,这样G++和C++都AC了:
int getval()
{       int temp1, temp2;
	switch(s[c_s++])
	{
	case 'p': return (value & (1 << 0))? 1:0;
	case 'q': return (value & (1 << 1))? 1:0;
	case 'r': return (value & (1 << 2))? 1:0;
	case 's': return (value & (1 << 3))? 1:0;
	case 't': return (value & (1 << 4))? 1:0;

	case 'K': temp1 = getval(); temp2 = getval(); return temp1 & temp2;
	case 'A': temp1 = getval(); temp2 = getval(); return temp1 | temp2;
	case 'N': return !getval();
	case 'C': temp1  = !getval(); temp2 = getval(); return temp1 | temp2;
	case 'E': temp1 = getval(); temp2 = getval(); return temp1 == temp2;
	}
}

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