Printing the data in the right format

Hi Guys
I am reading some array records from a device, i know the first 4 bytes i am reading in is the Calender date and time, how can i print this in a time format? Any help will be highly appreciated

Regards

Well, you have to know how the data is being stored. Number of seconds since some arbitrary start time? Anyway, check out time.h.

> Anyway, check out time.h.
http://www.cppreference.com/stddate/index.html

Quote:

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Originally Posted by dwks

> Anyway, check out time.h.
http://www.cppreference.com/stddate/index.html

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Hi There
All i know is that the first 4 bytes contain date and time information, Is there a way that i use this information to convert it into date and time by using some of the predefined functions in the above mentioned website.

Regards

There are too many ways that those four bytes could represent the date and time. Do you know the data format?

If not, you'll just have to guess. Here's my first guess:

Code:

---

time_t t;
FILE *fp;
fread(&t, sizeof t, 1, fp);
puts(ctime(&t));

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Quote:

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Originally Posted by dwks

There are too many ways that those four bytes could represent the date and time. Do you know the data format?

If not, you'll just have to guess. Here's my first guess:

Code:

---

time_t t;
FILE *fp;
fread(&t, sizeof t, 1, fp);
puts(ctime(&t));

---

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But this example will be readng from the file, i have my data saved in a data array, how would the above example be used to read the first four bytes from a array?

Well, you have to know how the data is represented in the array. How did you read it in, and what kind of thing did you store it in?

Code:

---

if (sizeof(time_t) != 4) printf("Huh? time_t is not 4 bytes\n");
else {
  time_t t;
  t = *(time_t *)&array[0];
}

---

This makes the assumption that the array location is such that a time_t object can be read directly from it. This is definitely OK on an x86 machine, but an array of char can be place at any byte boundary, and some processors will not be able to read a multibyte word from an unaligned address.

--
Mats

Quote:

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Originally Posted by matsp

Code:

---

if (sizeof(time_t) != 4) printf("Huh? time_t is not 4 bytes\n");
else {
  time_t t;
  t = *(time_t *)&array[0];
}

---

This makes the assumption that the array location is such that a time_t object can be read directly from it. This is definitely OK on an x86 machine, but an array of char can be place at any byte boundary, and some processors will not be able to read a multibyte word from an unaligned address.

--
Mats

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This is what i am tryin to do:

/*pnt2 is of type ubyte*/
pnt2 = &ans.data[8];
time_t t;
for(i=0;i<ans.data[2];i++)
{
if(array_type == 0) /* Analog array, each element is 4 bytes*/
{
ans_f.B[0] = pnt2[0];
ans_f.B[1] = pnt2[1];
ans_f.B[2] = pnt2[2];
ans_f.B[3] = pnt2[3];
pnt2 += 4;
t = ans_f.F;
printf("%s\n",ctime(&t));

This code is printing me the following thing

Wed Dec 31 17:00:00 1969

Does not seem right too me....i hope this helps you guys what i am trying to get.

Quote:

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Originally Posted by c_geek

This code is printing me the following thing

Wed Dec 31 17:00:00 1969

Does not seem right too me....i hope this helps you guys what i am trying to get.

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Sorry, I am not following... What do you think the output should be?
How it should the output look like?

Prefer you use code tags next time. If not, the board will strip all whitespace for your indentation and sometimes code is turned into smilies as well.

Quote:

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Originally Posted by c_geek

This is what i am tryin to do:

/*pnt2 is of type ubyte*/
pnt2 = &ans.data[8];
time_t t;
for(i=0;i<ans.data[2];i++)
{
if(array_type == 0) /* Analog array, each element is 4 bytes*/
{
ans_f.B[0] = pnt2[0];
ans_f.B[1] = pnt2[1];
ans_f.B[2] = pnt2[2];
ans_f.B[3] = pnt2[3];
pnt2 += 4;
t = ans_f.F;
printf("%s\n",ctime(&t));

This code is printing me the following thing

Wed Dec 31 17:00:00 1969

Does not seem right too me....i hope this helps you guys what i am trying to get.

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I'm guessing that means that t==0 (since that's what would be printed for a UNIX timestamp of 0 (midnight 1/1/70 GMT) on the east coast of the USA (GMT -5)).

Am I supposed to know what ans_f.F is and how it relates to ans_f.B? 'Cause I'm guessing that t is not set correctly.

Quote:

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Originally Posted by tabstop

I'm guessing that means that t==0 (since that's what would be printed for a UNIX timestamp of 0 (midnight 1/1/70 GMT) on the east coast of the USA (GMT -5)).

Am I supposed to know what ans_f.F is and how it relates to ans_f.B? 'Cause I'm guessing that t is not set correctly.

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Hi
i have defined a union so that i can collect the 4 bytes of data out of my ans.data. Here is the union struct:

Code:

---

union
{
  TLS_UBYTE B[4];                                /* No. of Bytes in Answer */
  TLS_FLOAT F;                                    /* Floats in Answer */
}ans_f

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Quote:

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Originally Posted by c_geek

Hi
i have defined a union so that i can collect the 4 bytes of data out of my ans.data. Here is the union struct:

Code:

---

union
{
  TLS_UBYTE B[4];                                /* No. of Bytes in Answer */
  TLS_FLOAT F;                                    /* Floats in Answer */
}ans_f

---

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It's possible that time_t is an integer, so that the assignment causes a cast and truncation of your data (interpreted as a float). Often floats are stored exponents first, so if your four-byte integer doesn't have any 1's in the first seven bits or so, you have a subnormal (read: very small) number.

It's definitely wrong to make it a float and then an integer [time_t is definitely not a float value].

--
Mats