Question

我正在尝试使用以下代码将字符串分解为整数和字符。在关于立即打印的第一部分中，我得到了正确的输出，但后来却出错了。

int Lottery::calcInvOdds(string ruleConstraint){
const char * sorted;
const char * unique;
string temp;
size_t pos;

temp = ruleConstraint;

pos = temp.find_first_of(" ");
sorted = temp.substr(0,pos).c_str();
cout << temp << endl;
cout << "S = " << sorted << endl;

 temp = temp.substr(pos+1);
 unique = temp.substr(0,pos).c_str();
 cout << "U = " << unique << endl;

cout << "Sorted = " << sorted[0] << " " << "Unique = " << unique[0] << endl<<endl;

return 0;
}

输出是这样的：

T F
S = T
U = F
Sorted = F Unique = F

F T
S = F
U = T
Sorted = T Unique = T

但是，在使用const char *和char sorted[2]等数组替换temp.substr(0,pos).c_str();并使用*temp.substr(0,pos).c_str()后，会显示正确的输出。这种行为的原因是什么？

Answer 1

sorted = temp.substr(0,pos).c_str();

这不会起作用。 temp.substr(0,pos)返回临时string，.c_str()获取指向其内容的指针，语句完成后释放临时string，使sorted指向释放存储器中。

您最好的选择是不要费心转换为const char*，而是将sorted和unique更改为string。然后事情会像你期望的那样工作，因为字符串将一直存在，直到函数退出。

int Lottery::calcInvOdds(const string& ruleConstraint){
    size_t pos = ruleConstraint.find_first_of(" ");
    string sorted = ruleConstraint.substr(0, pos);
    // The above line could be rewritten as:
    // string sorted(ruleConstraint, 0, pos);

    cout << ruleConstraint << endl;
    cout << "S = " << sorted << endl;

    // -- Not sure this is  what you want, but it's what your code does.
    #if 1
    string unique = ruleConstraint.substr(pos + 1, pos);

    // -- maybe you meant this
    #else
    size_t pos2 = ruleConstraint.find_first_of(" ", pos + 1);
    string unique(ruleConstraint, pos + 1, pos2 - pos - 1);
    #endif

    cout << "U = " << unique << endl;

    cout << "Sorted = " << sorted[0] << " " << "Unique = " << unique[0] << endl << endl;

    return 0;
}

const char *指向相同的内存位置

1 个答案: