Question

在使用VS 2010 / C ++交换两个变量的值的练习中，编译器“不”返回表达式的预期结果。

我理解一个更有效的表达式 - XOR x和y。本练习是使用代数表达式获得x和y的类似结果的另一种方法。

预期结果：x = 21，y = 11 样本CPP的结果是：x = 11，y = 11

// test.cpp : main project file.

#include "stdafx.h"

using namespace System;

int main(array<System::String ^> ^args)
{

    int x = 11; 
    int y = 23;

    x = (y+x) - (y = x);
    Console::WriteLine(System::String::Format("Value of x:", x));
    Console::WriteLine(System::String::Format("Value of y:", y));

    x = 11;
    y = 23;
    x = ((y+x)*1) - (y = x);
    Console::WriteLine(System::String::Format("Value of x:", x));
    Console::WriteLine(System::String::Format("Value of y:", y));

    return 0; 
}

ILDASM输出如下：

.method assembly static int32  main(string[] args) cil managed
{
 // Code size       118 (0x76)
   .maxstack  2
   .locals ([0] int32 y,
            [1] int32 x,
            [2] int32 V_2)
    IL_0000:  ldc.i4.0
    IL_0001:  stloc.2
    IL_0002:  ldc.i4.s   11
    IL_0004:  stloc.1
    IL_0005:  ldc.i4.s   23
    IL_0007:  stloc.0
    IL_0008:  ldloc.1
    IL_0009:  stloc.0
    IL_000a:  ldloc.0
    IL_000b:  ldloc.1
    IL_000c:  add
    IL_000d:  ldloc.0
    IL_000e:  sub
    IL_000f:  stloc.1
    IL_0010:  ldstr      "Value of x:"
    IL_0015:  ldloc.1
    IL_0016:  box        [mscorlib]System.Int32
    IL_001b:  call       string [mscorlib]System.String::Format(string,
                                                              object)
    IL_0020:  call       void [mscorlib]System.Console::WriteLine(string)
    IL_0025:  ldstr      "Value of y:"
    IL_002a:  ldloc.0
    IL_002b:  box        [mscorlib]System.Int32
    IL_0030:  call       string [mscorlib]System.String::Format(string,
                                                              object)
    IL_0035:  call       void [mscorlib]System.Console::WriteLine(string)
    IL_003a:  ldc.i4.s   11
    IL_003c:  stloc.1
    IL_003d:  ldc.i4.s   23
    IL_003f:  stloc.0
    IL_0040:  ldloc.1
    IL_0041:  stloc.0
    IL_0042:  ldloc.0
    IL_0043:  ldloc.1
    IL_0044:  add
    IL_0045:  ldloc.0
    IL_0046:  sub
    IL_0047:  stloc.1
    IL_0048:  ldstr      "Value of x:"
    IL_004d:  ldloc.1
    IL_004e:  box        [mscorlib]System.Int32
    IL_0053:  call       string [mscorlib]System.String::Format(string,
                                                              object)
    IL_0058:  call       void [mscorlib]System.Console::WriteLine(string)
    IL_005d:  ldstr      "Value of y:"
    IL_0062:  ldloc.0
    IL_0063:  box        [mscorlib]System.Int32
    IL_0068:  call       string [mscorlib]System.String::Format(string,
                                                              object)
    IL_006d:  call       void [mscorlib]System.Console::WriteLine(string)
    IL_0072:  ldc.i4.0
    IL_0073:  stloc.2
    IL_0074:  ldloc.2
    IL_0075:  ret
} // end of method 'Global Functions'::main

Answer 1

您的代码有未定义的行为。

x = (y+x) - (y = x);

在y分配中修改y = x以及在y中阅读y+x的值的副作用是无序的。你不能假设它们发生在另一个之前。除此之外，该标准明确指出这会导致不确定的行为。

参考：2011 ISO C ++标准，第1.9节[intro.execution]，第15段：

除非另有说明，否则评估各个运营商的操作数并且个别表达的子表达式没有被排序。 [...] 对运算符的操作数的值计算进行排序在运算符结果的值计算之前。如果一方对标量对象的影响相对于另一个对象没有统计对同一个标量对象的副作用或使用该值的计算值相同标量对象的值，行为未定义。

C ++运算符优先级（意外结果）

1 个答案: