Question

我得到了一个字符串y，我确保它只包含数字。在使用stoi函数将其存储在int变量中之前，如何检查它是否超出整数的边界？

string y = "2323298347293874928374927392374924"
int x = stoi(y); // The program gets aborted when I execute this as it exceeds the bounds
                 //   of int. How do I check the bounds before I store it?

Answer 1

您可以使用异常处理机制：

#include <stdexcept>

std::string y = "2323298347293874928374927392374924"
int x;

try {
  x = stoi(y);
}
catch(std::invalid_argument& e){
  // if no conversion could be performed
}
catch(std::out_of_range& e){
  // if the converted value would fall out of the range of the result type 
  // or if the underlying function (std::strtol or std::strtoull) sets errno 
  // to ERANGE.
}
catch(...) {
  // everything else
}

detailed description of stoi function and how to handle errors

Answer 2

抓住例外：

string y = "2323298347293874928374927392374924"
int x;

try {
  x = stoi(y);
}
catch(...) {
  // String could not be read properly as an int.
}

Answer 3

如果字符串表示一个太大而无法存储在int中的值，则将其转换为更大的值并检查结果是否适合int：

long long temp = stoll(y);
if (std::numeric_limits<int>::max() < temp
    || temp < std::numeric_limits<int>::min())
    throw my_invalid_input_exception();
int i = temp; // "helpful" compilers will warn here; ignore them.

在C ++中检查stoi（）函数中的int限制

3 个答案: