[英]How can I save a string input with blank space in c++ (I am using if else statement)
所以我正在尝试制作一个文本乘法器,这是代码
#include <iostream>
using namespace std;
int main()
{
bool m, n;
string x;
int y;
cout << "enter how many times you want to multiply the text : ";
cin >> y;
isdigit(y);
if (y)
{
cout << "enter the text you want to multiply : ";
cin >> x;
for (int a = 1; a <= y; ++a)
cout << x << endl;
}
else
{
cout << "you did not entered a number , try again";
}
return 0;
}
一切都很好,直到我知道它没有用空格保存文本输入我搜索了如何用空格存储字符串输入然后更改了代码但它没有用。 更改后的代码是
#include <iostream>
using namespace std;
int main()
{
bool m, n;
char x[100];
int y;
cout << "enter how many times you want to multiply the text : ";
cin >> y;
isdigit(y);
if (y)
{
cout << "enter the text you want to multiply : ";
cin.getline(x, 100);
for (int a = 1; a <= y; ++a)
cout << x << endl;
}
else
{
cout << "you did not entered a number , try again";
}
return 0;
}
请帮忙
如果我明白你想做什么,你需要读取 integer 值,清除operator>>
留在stdin
中的剩余'\n'
,然后使用getline()
读取你想要乘以的文本,例如
#include <iostream>
#include <limits>
using namespace std;
int main()
{
string x;
int y;
cout << "enter how many times you want to multiply the text : ";
if (!(cin >> y)) { /* validate stream-state after EVERY input */
std::cerr << "error: invalid integer input.\n";
return 1;
}
/* clear remaining '\n' from stdin (and any other characters) */
std::cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n');
cout << "enter the text you want to multiply : ";
if (!getline (cin, x)) { /* validate stream state */
std::cout << "user canceled input.\n";
return 0;
}
for (int a = 1; a <= y; ++a)
cout << x << endl;
return 0;
}
注意:使用isdigit(y)
是多余的。 如果您正确验证输入,则只需在读取后检查流状态即可确定在读取时是否输入了有效的 integer。 如果设置了失败位,则用户没有输入有效的failbit
。
虽然对于测试代码来说很好,但您需要查看为什么“使用命名空间标准;” 被认为是不好的做法?
示例使用/输出
$ ./bin/multiplytext
enter how many times you want to multiply the text : 3
enter the text you want to multiply : my dog has fleas
my dog has fleas
my dog has fleas
my dog has fleas
如果我误解了您的目标,请告诉我,我很乐意进一步提供帮助。
从这个答案可以看出,您正在混合>>
运算符和getline()
这会导致同步问题,因为getline
不等待输入刷新。
你可以打电话
cin.ignore();
或者
cin.clear();
cin.sync();
就在getline()
之前。
补丁代码:
#include <iostream>
using namespace std;
int main()
{
bool m, n;
char x[100];
int y;
cout << "enter how many times you want to multiply the text : ";
cin >> y;
isdigit(y);
if (y)
{
cout << "enter the text you want to multiply : ";
cin.ignore();
cin.getline(x, 100);
for (int a = 1; a <= y; ++a)
cout << x << endl;
}
else
{
cout << "you did not entered a number , try again";
}
return 0;
}
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