C++ 围绕变量"；变量名"；已损坏C++；

我有一个代码，应该是大学里一个项目的mips处理器的汇编程序。
现在我的问题是，当执行程序时，一切正常，直到程序退出主程序，它抛出以下错误：

变量

二进制指令

周围的堆栈已损坏

退出main时抛出错误，我在任何地方都找不到有效的解决方案

下面是我的代码示例：

#include <iostream>
#include <fstream>
#include <string>
#include <bitset>
using namespace std;
//function to convert string to binary
bool tobool(char x)
{
    if (x == '0')
        return 0;
    else
        return 1;
}
//function to decode the register name into an address using call by refrence to modify the instruction boolean encoding array
void register_encode(string Register, bool &address2, bool &address3, bool &address4, bool &address5, bool &address6)
{
    if (Register == "$zero")
    {
        address2 = 0;   address3 = 0;   address4 = 0;   address5 = 0;   address6 = 0;
    }
    else if (Register == "$t3")
    {
        address2 = 0; address3 = 1; address4 = 0; address5 = 1; address6 = 1;
    }
    else if (Register == "$s2")
    {
        address2 = 1; address3 = 0; address4 = 0; address5 = 1; address6 = 0;
    }
}

//function to count the number of occurence of commas (,) inside the string
int count_commas(string s) {
    int count = 0;

    for (int i = 0; i < s.size(); i++)
        if (s[i] == ',') count++;
    return count;
}

void main()
{
    int numberofinstruction = 0;
    string testingstring;
    string line[64];
    ifstream myfile;
    myfile.open("test.txt");
    // taking input from file into an array "line" which holds every single line inside the array in a single element inside the array
    while (getline(myfile, testingstring))
    {

        line[numberofinstruction] = testingstring;
        numberofinstruction++;
    }
    myfile.close();
    //transform to binary
    bool binaryinstruction[31];
    string firstterm, secondterm, thirdterm, fourthterm;
    for (int counter = 0; counter<numberofinstruction; counter++)
    {
        switch (count_commas(line[counter])) {
        case 1:
            firstterm = line[counter].substr(0, line[counter].find(" "));
            secondterm = line[counter].substr(line[counter].find(" ") + 1, line[counter].find(",") - line[counter].find(" ") - 1);
            thirdterm = line[counter].substr(line[counter].find(",") + 1, line[counter].size() - line[counter].find(","));
            if (firstterm == "lw")
            {   //LW encoding
                binaryinstruction[0] = 1;
                binaryinstruction[1] = 0;
                binaryinstruction[2] = 0;
                binaryinstruction[3] = 0;
                binaryinstruction[4] = 1;
                binaryinstruction[5] = 1;
                register_encode(secondterm, binaryinstruction[11], binaryinstruction[12], binaryinstruction[13], binaryinstruction[14], binaryinstruction[15]);
                string offset = thirdterm.substr(0, thirdterm.find('('));
                offset = bitset<16>(stoi(offset)).to_string();
                for (int x = 0; x <= 15; x++)
                {
                    binaryinstruction[16 + x] = tobool(offset[x]);
                }
                string lwregister = thirdterm.substr(thirdterm.find('$'), thirdterm.size() - thirdterm.find('(') - 2);
                register_encode(lwregister, binaryinstruction[6], binaryinstruction[7], binaryinstruction[8], binaryinstruction[9], binaryinstruction[10]);
            }
            else
            {   //SW encoding
                binaryinstruction[0] = 1;
                binaryinstruction[1] = 0;
                binaryinstruction[2] = 1;
                binaryinstruction[3] = 0;
                binaryinstruction[4] = 1;
                binaryinstruction[5] = 1;
                //begin
                register_encode(secondterm, binaryinstruction[11], binaryinstruction[12], binaryinstruction[13], binaryinstruction[14], binaryinstruction[15]);
                string offset = thirdterm.substr(0, thirdterm.find('('));
                offset = bitset<16>(stoi(offset)).to_string();
                for (int x = 0; x <= 15; x++)
                {
                    binaryinstruction[16 + x] = tobool(offset[x]);
                }
                string lwregister = thirdterm.substr(thirdterm.find('$'), thirdterm.size() - thirdterm.find('(') - 2);
                register_encode(lwregister, binaryinstruction[6], binaryinstruction[7], binaryinstruction[8], binaryinstruction[9], binaryinstruction[10]);
                //end

            }
            break;

        }
    }
    //testing
    ofstream myfile2("testout.txt");

    for (int f = 0; f <= 31; f = f + 8)
    {
        myfile2 << binaryinstruction[f] << binaryinstruction[f + 1] << binaryinstruction[f + 2] << binaryinstruction[f + 3] << binaryinstruction[f + 4] << binaryinstruction[f + 5] << binaryinstruction[f + 6] << binaryinstruction[f + 7] << endl;
    }
    //end testing
}

---

我的问题是访问不存在的数组的索引 我忘了，当在C++中识别一个数组时，你会键入数组中元素的数量，而不是数组最后的索引。

---

使我困惑的是错误发生的地方，当用断点调试时，它会退出异常，而退出Valuy主（）/<代码>，而访问未标识的数组元素时，不会发生异常。

< P>不像大多数其他高级语言，C++不一定执行数组索引检查。因此，索引错误会导致错误。在您的例子中，这会导致在退出时出现一条消息，但通常这种类型的错误完全没有报告

---

不过，有一些解决方案，特别是使用更现代的数组类型容器，如和。出于速度优化的原因，它们也不会在默认操作模式下执行索引检查，但是有一些方法可以启用它。请参阅我的答案。

这不是答案，但不能放在评论中

我们走吧

bool tobool(char x)
{
    if (x == '0')
        return 0;
    else
        return 1;

应该是

bool tobool(char x)
{
   return x == '0'? false : true;
}

void register_encode(const std::string& Register, bool &address2, bool &address3, bool &address4, bool &address5, bool &address6)

 int main()

然后

{

应该是

bool tobool(char x)
{
   return x == '0'? false : true;
}

void register_encode(const std::string& Register, bool &address2, bool &address3, bool &address4, bool &address5, bool &address6)

 int main()

{

同上

count_commas

我确信在

std:：string

中有一种方法可以做到这一点

也

应该是

bool tobool(char x)
{
   return x == '0'? false : true;
}

void register_encode(const std::string& Register, bool &address2, bool &address3, bool &address4, bool &address5, bool &address6)

 int main()

---

请提供一个testcase（代码示例），以便社区诊断实际问题。编写超出范围。如果不提供代码，您还想知道什么？@molbdnilo您可以指定具体位置吗？molbdnilo的意思是

bool二进制指令[31]；

有31个元素，索引从0到30（不是31！）。然后，您有

二进制指令[16+x]=tobool（偏移量[x]）；

其中x达到15，这与（int x＝0；x<代码>空main < /c> >是无效的C++。如果编译器没有告诉你，它还没有告诉你什么？尝试启用警告，而不要忽略警告。这不应该是一个答案，而是对问题的编辑。