如果cout丢失,循环进入无限循环

Question

我遇到了一些非常奇怪的事情.我遇到问题的代码是:

int stringPos;
int found1;
while (stringPos < 1);
{
    //start searching inString for framen starting at foundn and record
    found1 = inString.find(frame1, found1);
    cout << found1 << endl;


    //if return is a number, push back foundn to a vector
    if (found1 != -1)
    {
        foundPositions.push_back(found1);
    }
    //if return is npos, then break the loop
    else
    {
        stringPos=1;
    }

    //add 1 to foundn so that the search would continue from where the
    //search ended last
    found1+=1;
}

奇怪的是,当我放在线cout << found1 << endl;下面时found1 = inString.find(frame1, found1);,循环正确执行.但是,如果我没有cout << found1 << endl;它进入无限循环......

有什么建议？谢谢!

Answer 1

这是一个错误(并使用一个整数变量):

while (stringPos < 1);

因为它相当于:

while (stringPos < 1) {}

如果没有进入无限循环,则跟随它的代码将仅执行一次.纠正:

• 初始化变量stringPos和found1.
• 使用类型size_t为stringPos和found作为std::string::find()不返回int,但回报率size_type(一般size_t).
• 使用std::string::npos而不是-1测试未找到.
• 删除尾部分号.