c++ - 确定 strftime char 数组的最大大小的智能方法是什么？

Question

如何在不反复试验的情况下为 strftime 调整 char 数组的大小？使用 mktime，示例中的时间戳大小 N 必须大于 86，否则我会返回任意日期。 例如

N = 86:2013-07-13 02:41

N = 82:1979-05-18 13:23

如何在事先不知道日期的情况下有效地缩放 N？检查 >0 没有帮助。

#include <iostream>
#include <cstring>
#include <ctime>

#define N 86

using namespace std;

int main(void)
{
time_t t;
struct tm ts;
char timestamp[N] ;

ts.tm_min    = 41;     
ts.tm_hour   = 2;     
ts.tm_mday   = 13;
ts.tm_mon    = 7 - 1;      
ts.tm_year   = 13 - 1900 + 2000;         

t = mktime(&ts);

if (strftime(timestamp, sizeof(timestamp)-1, "%Y-%m-%d %H:%M", &ts) > 0)
    cout << timestamp;
else {
    cerr << "strftime failed." <<endl;
    return 1;
}

return 0;
}

Answer 1

来自 strftime 的文档:

If the length of the resulting C string, including the terminating null-character, doesn't exceed maxsize, the function returns the total number of characters copied to ptr (not including the terminating null-character). Otherwise, it returns zero, and the contents of the array pointed by ptr are indeterminate.

这意味着如果您不知道大小并且可以动态分配一个字符串，您可以按照以下方式做一些事情:

int size = N; //  Some starting size
char *timestamp = malloc(size);

//  Your time stuff

int result = strftime(timestamp, size - 1, "%Y-%m-%d %H:%M", &ts);

// While there isn't enough room to store the result
while (result == 0)
{
  free(timestamp);  //  Free old data
  size *= 2;        //  Double the size (should be more than enough)
  timestamp = malloc(size); //  Allocate the new size. You can check for failed allocations here as well.

  //  Retry
  result = strftime(timestamp, size - 1, "%Y-%m-%d %H:%M", &ts);
}

std::cout << timestamp;