Date in mm-dd or similar

I'm trying to have a string that is the date, preferably in mm-dd format, though something like dd-mm-yyyy is equally suitable. I found out how to do it in C, and I found examples in C++ that were a MyFunction.print function recall, and I couldn't make them into a string.

>> I'm trying to have a string that is the date, preferably in mm-dd format, though something like dd-mm-yyyy is equally suitable. I found out how to do it in C, and I found examples in C++ that were a MyFunction.print function recall, and I couldn't make them into a string.

Why don't you just use a C function like ctime and assign it to an std::string?

Good idea.

Code:

---

#include <stdio.h>
#include <time.h>

int main()
{
time_t now;
struct tm *d;
char li[13];

time(&now);
d = localtime(&now);

strftime(li, 15, "%d-%m-%Y", d);

printf("%s\n", li);

return 0;
}

---

There's a program in C that does what I want. However, I don't know what to do, other than changing the #includes and printf and soforth.

That's about all that needs to be done. Using std::string is not that different.

string li = "mm-dd-yyyy"; // filler string

strftime( li.data(), li.size(), "%d-%m-%Y", d);

Code:

---

#include <iostream>
#include <ctime>
#include <string>

using namespace std;

int main()
{
time_t now;
struct tm *d;
string li = "mm-dd-yyyy"; // filler string

time(&now);
d = localtime(&now);

strftime( li.data(), li.size(), "%d-%m-%Y", d);

cout << s, li;

}

---

./ctime_conv.cpp: In function ‘int main()’:
./ctime_conv.cpp:16: error: invalid conversion from ‘const char*’ to ‘char*’
./ctime_conv.cpp:16: error: initializing argument 1 of ‘size_t strftime(char*, size_t, const char*, const tm*)’
./ctime_conv.cpp:18: error: ‘s’ was not declared in this scope
./ctime_conv.cpp:18: error: expected `;' before ‘li’

>> ./ctime_conv.cpp:16: error: invalid conversion from ‘const char*’ to ‘char*’

Cast the return value of the data( ) function to char*.

>> cout << s, li;

Typo?

Get rid of li or s?

>> Get rid of li or s?

Where is 's' even declared? And what operators does ostream define? Left shift, yes. But what others?

Ok, to make things simpler, here's the C source.

Code:

---

#include <stdio.h>
#include <time.h>

int main()
{
time_t now;
struct tm *d;
char li[13];

time(&now);
d = localtime(&now);

strftime(li, 15, "%d-%m-%Y", d);

printf("%s\n", li);

return 0;
}

---

It outputs 17-07-2009.

>> Ok, to make things simpler, here's the C source.

Yes, we've already seen that bit of code. What has that got to do with the code you posted with errors?

The program I built is in C++, and this is in C. I don't know how to change this to C++.

Quote:

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Originally Posted by Muscovy

The program I built is in C++, and this is in C. I don't know how to change this to C++.

---

Well you got pretty far on your own, just read your errors.

Quote:

---

Originally Posted by Muscovy

./ctime_conv.cpp: In function ‘int main()’:
./ctime_conv.cpp:16: error: invalid conversion from ‘const char*’ to ‘char*’
./ctime_conv.cpp:16: error: initializing argument 1 of ‘size_t strftime(char*, size_t, const char*, const tm*)’
./ctime_conv.cpp:18: error: ‘s’ was not declared in this scope
./ctime_conv.cpp:18: error: expected `;' before ‘li’

---

The error on line 16 is from the data( ) function when I showed you how to call strftime. You need to cast the return result to a char * type so the function can write the new date string in li. Sorry I didn't do that earlier.

The error on line 18 has to do with the cout statement you wrote. "s was not declared" should be clear to a C or C++ person, but I don't see an s variable in your code. Printing li should work if you delete ", s" in your code.

Then you recompile, work through your errors like this again, and so forth until you succeed.

YES! It works!