[英]What is an intelligent way to determine max size of a strftime char array?
How can I size the char array for strftime without trial and error? 如何在没有尝试和错误的情况下为strftime确定char数组的大小? Using mktime, the timestamp size N in the example has to be greater 86, otherwise I get arbitrary dates back.
使用mktime,示例中的时间戳大小N必须大于86,否则我会得到任意日期。 eg
例如
N = 86 : 2013-07-13 02:41
N = 86 :
2013-07-13 02:41
N = 82 : 1979-05-18 13:23
N = 82 :
1979-05-18 13:23
How do I efficiently scale N without prior knowledge of the date? 如何在不事先知道日期的情况下有效地缩放N? The check >0 does not help.
大于0的检查无济于事。
#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;
}
From the documentation for strftime: 从strftime的文档中:
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).
如果生成的C字符串的长度(包括终止的空字符)不超过maxsize,则该函数返回复制到ptr的字符总数(不包括终止的空字符)。 Otherwise, it returns zero, and the contents of the array pointed by ptr are indeterminate.
否则,它返回零,并且ptr指向的数组的内容是不确定的。
That means if you don't know the size and can dynamically allocate a string you can do something along the lines of: 这意味着,如果您不知道大小,并且可以动态分配字符串,则可以执行以下操作:
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;
Because you tagged this as C++, perhaps you might consider the following. 因为您将其标记为C ++,所以也许您可以考虑以下内容。
--> Note that there is no struggle with the string size here. ->请注意,这里的字符串大小没有任何困难。
// function to create a timestamp style string
std::string yyDmmDdd_hhCmmGet(time_t tt)
{
std::stringstream ss;
// goal - something like: "%Y-%m-%d %H:%M"
{
struct tm mybdtod; // linux api: my broken down time of day
// the following is a relatively slow function
::localtime_r (&tt, &mybdtod);
// linux api - convert time_t to tm as local time
ss << std::setw(4) << (mybdtod.tm_year+1900)
<< "-"
<< std::setfill('0') << std::setw(2) << mybdtod.tm_mon+1
<< "-"
<< std::setfill('0') << std::setw(2) << mybdtod.tm_mday
<< " ";
ss << std::dec << std::setfill('0') << std::setw(2)
<< mybdtod.tm_hour
<< ":"
<< std::setfill('0') << std::setw(2)
<< mybdtod.tm_min;
}
return(ss.str());
}
int t186(void)
{
struct tm ts; // linux api: time struct
::memset(&ts, 0, sizeof(tm));
ts.tm_min = 41;
ts.tm_hour = 3-1;
ts.tm_mday = 13;
ts.tm_mon = 7 - 1;
ts.tm_year = 13 - 1900 + 2000;
time_t tt = mktime(&ts); // linux api: Convert tm struct to time_t
// format time_t to string
std::string s = yyDmmDdd_hhCmmGet(tt); // timestamp style
std::cout << "\n" << s
<< "\n s.size(): "
<< s.size() << " chars" << std::endl;
// now we know how many chars timestamp needs
// add 1 to size because ?strftime seems to need it?
char timestamp[s.size()+1];
(void)strftime(timestamp, sizeof(timestamp), "%Y-%m-%d %H:%M", &ts);
// linux api: format time_t to string
std::cout << "\n" << timestamp << std::endl;
std::cout << " sizeof(timestamp): "
<< sizeof(timestamp) << " chars" << std::endl;
return(0);
}
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