Question

在我的程序中，我有一个时钟计时器，我需要将它保存到一个char数组[10]，然后将其实现到我的高分函数中。通过我的程序，我已经有一个格式化的时间。例如，如果时钟的秒数小于10，我必须加零。所以，0：02，如果时钟的秒数大于10，则保持正常。而不是我在我的struct中使用两个int变量，我怎么才能在我的文本文件中写一个字符串？例如，让我们写一个名为string clocksTime =“0:15”的字符串。请注意，它已被格式化。这是我的代码：

struct highscore
{
    // *Want* Change these two int values to a string
    int clockMin;
    int clockSec;
}; 
...[Code]…
// Change the two variables to have it instead, data[playerScore].clock = clocksTime
data[playerScore].clockMin = clockData.minutes;
data[playerScore].clockSec = clockData.seconds;
_strdate(data[playerScore].Date);

// Write the variables into the text file
streaming = fopen( "Highscores.dat", "wb" );
fwrite( data, sizeof(data), 1 , streaming);

Answer 1

If you can use "\\\\\\1" it's a lot cleaner. Here is a program that does it:

std::string

If, on the other hand, you can't use #include <iostream> using namespace std; void int_time_to_string(const int int_time, string& string_time); int main() { int clockMin = 0; int clockSec = 15; string clockMinString; string clockSecString; int_time_to_string(clockMin, clockMinString); int_time_to_string(clockSec, clockSecString); string clockString = clockMinString + ":" + clockSecString; cout << "clock = " << clockString << endl; return 0; } // Convert a time in int to char*, and pad with a 0 if there's only one digit void int_time_to_string(const int int_time, string& string_time) { if (int_time < 10) { // Only one digit, pad with a 0 string_time = "0" + to_string(int_time); } else { // Two digits, it's already OK string_time = to_string(int_time); } }, you have to resort to C strings, that is, arrays of chars. I'm quite rusty about this, but I think we can use std::string for the conversion, and sprintf/strcpy to deal with some of the lower-level operations (like putting '\0' at the end).

strcat

In both cases you just have to take #include <iostream> #include <cstring> using namespace std; void int_time_to_string(const int int_time, char* string_time); int main() { int clockMin = 0; int clockSec = 15; char clockMinString[3] = ""; // 2 digits + null character char clockSecString[3] = ""; // 2 digits + null character int_time_to_string(clockMin, clockMinString); int_time_to_string(clockSec, clockSecString); char clockString[6]; // 2 digits, colon, 2 more digits, null character strcpy(clockString, clockMinString); strcat(clockString, ":"); strcat(clockString, clockSecString); cout << "clock = " << clockString << endl; return 0; } // Convert a time in int to char*, and pad with a 0 if there's only one digit void int_time_to_string(const int int_time, char* string_time) { if (int_time < 10) { // Only one digit, pad with a 0 strcpy(string_time, "0") ; char temp[2]; // 1 digit + null character sprintf(temp, "%d", int_time); strcat(string_time, temp); } else { // Two digits, it's already OK sprintf(string_time, "%d", int_time); } }.

C ++将结构中的字符串保存到文本文件中

1 个答案: