我正在写一个函数来返回一个数字的倒数,也就是说,它将int(1234)转换为int(4321)。这是我目前所拥有的:
#include <iostream>
#include <cstdlib>
#include <string>
#include <sstream>
using namespace std;
int reverse(int num) {
stringstream ss (stringstream::in | stringstream::out);
string initial;
int reversed;
// read the number in to a string stream
ss << num;
initial = ss.str();
// flush the stringstream
ss.str("");
for(unsigned int i(0); i <= initial.size(); i++) {
ss << initial[initial.size() - i];
}
ss >> reversed;
return reversed;
}
int main(int argc, const char *argv[])
{
int test = 9871;
cout << "test = " << test << endl;
cout << "reverse = " << reverse(test) << endl;
return 0;
}然而,这只是输出:
test = 9871
reverse = 0我非常确定问题出在ss >> reversed这一行,问题出在reversed被设置为0而不是ss的值,但是我不知道这段代码有什么问题,它看起来应该很简单,这让人恼火。有人能帮上忙吗?
谢谢
发布于 2012-05-30 23:17:44
i从0开始,然后initial.size() - i超出字符串界限。
更改为ss << initial[initial.size() - i - 1];并将i从1迭代为initiali.size() - 1
for(unsigned i = 0; i != initial.size(); ++i) {
ss << initial[initial.size() - i -1];
}或者从1迭代i到initial.size()
for(unsigned i = 1; i <= initial.size(); ++i) {
ss << initial[initial.size() - i];
}发布于 2012-05-30 23:21:16
for循环会导致对initial的越界访问。除了Alessandro Pezzato指出的问题外,for循环中的终止条件需要更改为i < initial.size(),否则仍然会发生越界:
for(unsigned int i(0); i < initial.size(); i++) {
ss << initial[initial.size() - i - 1];
}发布于 2012-05-30 23:21:46
执行所需操作的最快C++11方法是:
string s = std::to_string(my_int);
std::reverse(begin(s), end(s));
return std::stoi(s);https://stackoverflow.com/questions/10819161
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