Using gdb to print a time_t variable-CodePudding

I want to print some information from gdb but don't see how. I am used to p/s p/x formats. But don't know what to do in the case below.

#include<iostream>
#include<climits>
#include <stdio.h>
#include <time.h>
#include <stdint.h>

using namespace std;
int main()
{
        time_t dataFrom = 1234560;
        cout << "dataFrom = " << asctime(gmtime(&dataFrom)) << "\n";

        return 0;
}

How do i print dataFrom using gdb correctly ? I have read about informing gdb about a structure if it was user defined. But what to do with something inbuilt such as time_t ?

badri@badri-All-Series:~/progs$ g   --std=c  11 testgdb.cpp -g
badri@badri-All-Series:~/progs$ gdb ./a.out 
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Reading symbols from ./a.out...done.
(gdb) b main
Breakpoint 1 at 0x903: file testgdb.cpp, line 9.
(gdb) r
Starting program: /home/badri/progs/a.out 

Breakpoint 1, main () at testgdb.cpp:9
9   {
(gdb) n
10      time_t dataFrom = 1234560;
(gdb) p dataFrom
$1 = 93824992233952
(gdb) p *dataFrom
$2 = 1447122753
(gdb) p/s dataFrom
$3 = 93824992233952

CodePudding user response：

Yes, as mentioned by n. 1.8e9-where's-my-share m, the execution is stopped at the line 9 and the line 10 is yet to be executed. So, a simple next or n command in gdb will run the line 10 and then if you print the value in dataFrom you can observe the proper value.

GNU gdb (Ubuntu 9.2-0ubuntu1~20.04) 9.2
...
Reading symbols from ./a.out...
(gdb) b main
Breakpoint 1 at 0x11e9: file testgdb.cpp, line 9.
(gdb) r
Starting program: /home/user/path/a.out

Breakpoint 1, main () at testgdb.cpp:9
9       {
(gdb) n
10              time_t dataFrom = 1234560;
(gdb) p dataFrom
$1 = 0
(gdb) n
11              cout << "dataFrom = " << asctime(gmtime(&dataFrom)) << "\n";
(gdb) p dataFrom
$2 = 1234560
(gdb)