C Program Time Conversion

I came upon this Time Conversion Program in HackerRank, I was surprised the way how the program is compiled in HackerRank (or could be my misunderstanding of C).

#include <stdio.h>



int main() {

    int hh, mm, ss;

    char tt[2];

    scanf("%d:%d:%d%s", &hh, &mm, &ss, tt);

    printf("%dn", hh);

    if(strcmp(tt, "PM") == 0 && hh != 12) hh += 12;

    if(strcmp(tt, "AM") == 0 && hh == 12) hh = 0;

    printf("%02d:%02d:%02d", hh, mm, ss);

    return 0;

}

The above program, when I ran in my computer, using MinGW 32-bit GCC Compiler, I was getting the value of hh as zero.

enter image description here

Fine, I thought could be compiler issue and ran the code in IDEONE, same results.

enter image description here

But when I ran this same code with HackerRank, all testcases were passed, I have no idea, how this is working here?

enter image description here

I am confused at this stage, am I doing this right?

edited Aug 11 '17 at 20:14

Martijn Pieters♦

724k14325452348

asked Apr 12 '16 at 10:27

cs1193

84611222

3

You are 1-char short on char tt[2]; You forgot about the nul-terminating character. You need at minimum char tt[3] = ""; to hold AM or PM (the initialization is just good practice)

– David C. Rankin
Apr 12 '16 at 10:32

@DavidC.Rankin, how does variable hh return zero, just because of char pointer is having a shorter storage space? and how does it run on hackerrank?

– cs1193
Apr 12 '16 at 10:35

3

It runs poorly. You invoke undefined behavior.

– Lundin
Apr 12 '16 at 10:41

2

@cs1193 It depends on the layout of variables on the stack. If tt is right before hh, the final nul character will overwrite the first byte of hh, setting it to zero on a little endian architecture.

– nwellnhof
Apr 12 '16 at 11:01

add a comment |

I came upon this Time Conversion Program in HackerRank, I was surprised the way how the program is compiled in HackerRank (or could be my misunderstanding of C).

#include <stdio.h>



int main() {

    int hh, mm, ss;

    char tt[2];

    scanf("%d:%d:%d%s", &hh, &mm, &ss, tt);

    printf("%dn", hh);

    if(strcmp(tt, "PM") == 0 && hh != 12) hh += 12;

    if(strcmp(tt, "AM") == 0 && hh == 12) hh = 0;

    printf("%02d:%02d:%02d", hh, mm, ss);

    return 0;

}

The above program, when I ran in my computer, using MinGW 32-bit GCC Compiler, I was getting the value of hh as zero.

enter image description here

Fine, I thought could be compiler issue and ran the code in IDEONE, same results.

enter image description here

But when I ran this same code with HackerRank, all testcases were passed, I have no idea, how this is working here?

enter image description here

I am confused at this stage, am I doing this right?

edited Aug 11 '17 at 20:14

Martijn Pieters♦

724k14325452348

asked Apr 12 '16 at 10:27

cs1193

84611222

3

You are 1-char short on char tt[2]; You forgot about the nul-terminating character. You need at minimum char tt[3] = ""; to hold AM or PM (the initialization is just good practice)

– David C. Rankin
Apr 12 '16 at 10:32

@DavidC.Rankin, how does variable hh return zero, just because of char pointer is having a shorter storage space? and how does it run on hackerrank?

– cs1193
Apr 12 '16 at 10:35

3

It runs poorly. You invoke undefined behavior.

– Lundin
Apr 12 '16 at 10:41

2

@cs1193 It depends on the layout of variables on the stack. If tt is right before hh, the final nul character will overwrite the first byte of hh, setting it to zero on a little endian architecture.

– nwellnhof
Apr 12 '16 at 11:01

add a comment |

I came upon this Time Conversion Program in HackerRank, I was surprised the way how the program is compiled in HackerRank (or could be my misunderstanding of C).

#include <stdio.h>



int main() {

    int hh, mm, ss;

    char tt[2];

    scanf("%d:%d:%d%s", &hh, &mm, &ss, tt);

    printf("%dn", hh);

    if(strcmp(tt, "PM") == 0 && hh != 12) hh += 12;

    if(strcmp(tt, "AM") == 0 && hh == 12) hh = 0;

    printf("%02d:%02d:%02d", hh, mm, ss);

    return 0;

}

The above program, when I ran in my computer, using MinGW 32-bit GCC Compiler, I was getting the value of hh as zero.

enter image description here

Fine, I thought could be compiler issue and ran the code in IDEONE, same results.

enter image description here

But when I ran this same code with HackerRank, all testcases were passed, I have no idea, how this is working here?

enter image description here

I am confused at this stage, am I doing this right?

edited Aug 11 '17 at 20:14

Martijn Pieters♦

724k14325452348

asked Apr 12 '16 at 10:27

cs1193

84611222

I came upon this Time Conversion Program in HackerRank, I was surprised the way how the program is compiled in HackerRank (or could be my misunderstanding of C).

#include <stdio.h>



int main() {

    int hh, mm, ss;

    char tt[2];

    scanf("%d:%d:%d%s", &hh, &mm, &ss, tt);

    printf("%dn", hh);

    if(strcmp(tt, "PM") == 0 && hh != 12) hh += 12;

    if(strcmp(tt, "AM") == 0 && hh == 12) hh = 0;

    printf("%02d:%02d:%02d", hh, mm, ss);

    return 0;

}

The above program, when I ran in my computer, using MinGW 32-bit GCC Compiler, I was getting the value of hh as zero.

enter image description here

Fine, I thought could be compiler issue and ran the code in IDEONE, same results.

enter image description here

But when I ran this same code with HackerRank, all testcases were passed, I have no idea, how this is working here?

enter image description here

I am confused at this stage, am I doing this right?

c time

edited Aug 11 '17 at 20:14

Martijn Pieters♦

724k14325452348

asked Apr 12 '16 at 10:27

cs1193

84611222

edited Aug 11 '17 at 20:14

Martijn Pieters♦

724k14325452348

asked Apr 12 '16 at 10:27

cs1193

84611222

edited Aug 11 '17 at 20:14

Martijn Pieters♦

724k14325452348

edited Aug 11 '17 at 20:14

Martijn Pieters♦

724k14325452348

edited Aug 11 '17 at 20:14

Martijn Pieters♦

724k14325452348

asked Apr 12 '16 at 10:27

cs1193

84611222

asked Apr 12 '16 at 10:27

cs1193

84611222

asked Apr 12 '16 at 10:27

cs1193

84611222

3

You are 1-char short on char tt[2]; You forgot about the nul-terminating character. You need at minimum char tt[3] = ""; to hold AM or PM (the initialization is just good practice)

– David C. Rankin
Apr 12 '16 at 10:32

@DavidC.Rankin, how does variable hh return zero, just because of char pointer is having a shorter storage space? and how does it run on hackerrank?

– cs1193
Apr 12 '16 at 10:35

3

It runs poorly. You invoke undefined behavior.

– Lundin
Apr 12 '16 at 10:41

2

@cs1193 It depends on the layout of variables on the stack. If tt is right before hh, the final nul character will overwrite the first byte of hh, setting it to zero on a little endian architecture.

– nwellnhof
Apr 12 '16 at 11:01

add a comment |

3

You are 1-char short on char tt[2]; You forgot about the nul-terminating character. You need at minimum char tt[3] = ""; to hold AM or PM (the initialization is just good practice)

– David C. Rankin
Apr 12 '16 at 10:32

@DavidC.Rankin, how does variable hh return zero, just because of char pointer is having a shorter storage space? and how does it run on hackerrank?

– cs1193
Apr 12 '16 at 10:35

3

It runs poorly. You invoke undefined behavior.

– Lundin
Apr 12 '16 at 10:41

2

@cs1193 It depends on the layout of variables on the stack. If tt is right before hh, the final nul character will overwrite the first byte of hh, setting it to zero on a little endian architecture.

– nwellnhof
Apr 12 '16 at 11:01

You are 1-char short on char tt[2]; You forgot about the nul-terminating character. You need at minimum char tt[3] = ""; to hold AM or PM (the initialization is just good practice)

– David C. Rankin
Apr 12 '16 at 10:32

@DavidC.Rankin, how does variable hh return zero, just because of char pointer is having a shorter storage space? and how does it run on hackerrank?

– cs1193
Apr 12 '16 at 10:35

It runs poorly. You invoke undefined behavior.

– Lundin
Apr 12 '16 at 10:41

@cs1193 It depends on the layout of variables on the stack. If tt is right before hh, the final nul character will overwrite the first byte of hh, setting it to zero on a little endian architecture.

– nwellnhof
Apr 12 '16 at 11:01

add a comment |

1 Answer
1

active

oldest

votes

With the changes made to allow space in tt to accommodate the nul-terminating character, the code functions well in making conversions to military time. By initializing tt to zero all elements during declaration/definition, you have insure tt will be nul-terminated when AM or PM are added. e.g.

#include <stdio.h>

#include <string.h>



int main (void) {



    int hh, mm, ss;

    char tt[3] = "";



    printf (" enter time in HH:MM:SS AM/PM: ");

    if (scanf ("%d:%d:%d %[^n]%*c", &hh, &mm, &ss, tt) != 4) {

        fprintf (stderr, "error: invalid read of time.n");

    }



    hh = (strcmp (tt, "PM") == 0 && hh != 12) ? hh + 12 : hh;

    if (strcmp (tt, "AM") == 0 && hh == 12) hh = 0;

    printf (" %02d:%02d:%02dnn", hh, mm, ss);



    return 0;

}

Example Use/Output

$ ./bin/tmread

 enter time in HH:MM:SS AM/PM: 11:59:04 PM

 23:59:04

Look it over and let me know if you have any questions.

edited Apr 14 '16 at 16:12

answered Apr 13 '16 at 8:08

David C. Rankin

43.3k33151

scanf with "%s" is dangerous, since it does not limit the string length. See here msdn.microsoft.com/en-us/library/xdb9w69d.aspx or other places for how to avoid this problem in general.

– Stian Skjelstad
Apr 14 '16 at 10:50

Agreed, it can always be replaced with "%[^n]%*c" which is probably the better solution. It was originally left as %s since we were dealing with AM, PM in a fixed format file, it avoided having to discuss the use of the assignment suppression operator, etc...

– David C. Rankin
Apr 14 '16 at 16:11

add a comment |

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1 Answer
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1 Answer
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votes

#include <stdio.h>

#include <string.h>



int main (void) {



    int hh, mm, ss;

    char tt[3] = "";



    printf (" enter time in HH:MM:SS AM/PM: ");

    if (scanf ("%d:%d:%d %[^n]%*c", &hh, &mm, &ss, tt) != 4) {

        fprintf (stderr, "error: invalid read of time.n");

    }



    hh = (strcmp (tt, "PM") == 0 && hh != 12) ? hh + 12 : hh;

    if (strcmp (tt, "AM") == 0 && hh == 12) hh = 0;

    printf (" %02d:%02d:%02dnn", hh, mm, ss);



    return 0;

}

Example Use/Output

$ ./bin/tmread

 enter time in HH:MM:SS AM/PM: 11:59:04 PM

 23:59:04

Look it over and let me know if you have any questions.

edited Apr 14 '16 at 16:12

answered Apr 13 '16 at 8:08

David C. Rankin

43.3k33151

scanf with "%s" is dangerous, since it does not limit the string length. See here msdn.microsoft.com/en-us/library/xdb9w69d.aspx or other places for how to avoid this problem in general.

– Stian Skjelstad
Apr 14 '16 at 10:50

Agreed, it can always be replaced with "%[^n]%*c" which is probably the better solution. It was originally left as %s since we were dealing with AM, PM in a fixed format file, it avoided having to discuss the use of the assignment suppression operator, etc...

– David C. Rankin
Apr 14 '16 at 16:11

add a comment |

#include <stdio.h>

#include <string.h>



int main (void) {



    int hh, mm, ss;

    char tt[3] = "";



    printf (" enter time in HH:MM:SS AM/PM: ");

    if (scanf ("%d:%d:%d %[^n]%*c", &hh, &mm, &ss, tt) != 4) {

        fprintf (stderr, "error: invalid read of time.n");

    }



    hh = (strcmp (tt, "PM") == 0 && hh != 12) ? hh + 12 : hh;

    if (strcmp (tt, "AM") == 0 && hh == 12) hh = 0;

    printf (" %02d:%02d:%02dnn", hh, mm, ss);



    return 0;

}

Example Use/Output

$ ./bin/tmread

 enter time in HH:MM:SS AM/PM: 11:59:04 PM

 23:59:04

Look it over and let me know if you have any questions.

edited Apr 14 '16 at 16:12

answered Apr 13 '16 at 8:08

David C. Rankin

43.3k33151

scanf with "%s" is dangerous, since it does not limit the string length. See here msdn.microsoft.com/en-us/library/xdb9w69d.aspx or other places for how to avoid this problem in general.

– Stian Skjelstad
Apr 14 '16 at 10:50

Agreed, it can always be replaced with "%[^n]%*c" which is probably the better solution. It was originally left as %s since we were dealing with AM, PM in a fixed format file, it avoided having to discuss the use of the assignment suppression operator, etc...

– David C. Rankin
Apr 14 '16 at 16:11

add a comment |

#include <stdio.h>

#include <string.h>



int main (void) {



    int hh, mm, ss;

    char tt[3] = "";



    printf (" enter time in HH:MM:SS AM/PM: ");

    if (scanf ("%d:%d:%d %[^n]%*c", &hh, &mm, &ss, tt) != 4) {

        fprintf (stderr, "error: invalid read of time.n");

    }



    hh = (strcmp (tt, "PM") == 0 && hh != 12) ? hh + 12 : hh;

    if (strcmp (tt, "AM") == 0 && hh == 12) hh = 0;

    printf (" %02d:%02d:%02dnn", hh, mm, ss);



    return 0;

}

Example Use/Output

$ ./bin/tmread

 enter time in HH:MM:SS AM/PM: 11:59:04 PM

 23:59:04

Look it over and let me know if you have any questions.

edited Apr 14 '16 at 16:12

answered Apr 13 '16 at 8:08

David C. Rankin

43.3k33151

#include <stdio.h>

#include <string.h>



int main (void) {



    int hh, mm, ss;

    char tt[3] = "";



    printf (" enter time in HH:MM:SS AM/PM: ");

    if (scanf ("%d:%d:%d %[^n]%*c", &hh, &mm, &ss, tt) != 4) {

        fprintf (stderr, "error: invalid read of time.n");

    }



    hh = (strcmp (tt, "PM") == 0 && hh != 12) ? hh + 12 : hh;

    if (strcmp (tt, "AM") == 0 && hh == 12) hh = 0;

    printf (" %02d:%02d:%02dnn", hh, mm, ss);



    return 0;

}

Example Use/Output

$ ./bin/tmread

 enter time in HH:MM:SS AM/PM: 11:59:04 PM

 23:59:04

Look it over and let me know if you have any questions.

edited Apr 14 '16 at 16:12

answered Apr 13 '16 at 8:08

David C. Rankin

43.3k33151

edited Apr 14 '16 at 16:12

answered Apr 13 '16 at 8:08

David C. Rankin

43.3k33151

answered Apr 13 '16 at 8:08

David C. Rankin

43.3k33151

answered Apr 13 '16 at 8:08

David C. Rankin

43.3k33151

scanf with "%s" is dangerous, since it does not limit the string length. See here msdn.microsoft.com/en-us/library/xdb9w69d.aspx or other places for how to avoid this problem in general.

– Stian Skjelstad
Apr 14 '16 at 10:50

Agreed, it can always be replaced with "%[^n]%*c" which is probably the better solution. It was originally left as %s since we were dealing with AM, PM in a fixed format file, it avoided having to discuss the use of the assignment suppression operator, etc...

– David C. Rankin
Apr 14 '16 at 16:11

add a comment |

scanf with "%s" is dangerous, since it does not limit the string length. See here msdn.microsoft.com/en-us/library/xdb9w69d.aspx or other places for how to avoid this problem in general.

– Stian Skjelstad
Apr 14 '16 at 10:50

Agreed, it can always be replaced with "%[^n]%*c" which is probably the better solution. It was originally left as %s since we were dealing with AM, PM in a fixed format file, it avoided having to discuss the use of the assignment suppression operator, etc...

– David C. Rankin
Apr 14 '16 at 16:11

scanf with "%s" is dangerous, since it does not limit the string length. See here msdn.microsoft.com/en-us/library/xdb9w69d.aspx or other places for how to avoid this problem in general.

– Stian Skjelstad
Apr 14 '16 at 10:50

Agreed, it can always be replaced with "%[^n]%*c" which is probably the better solution. It was originally left as %s since we were dealing with AM, PM in a fixed format file, it avoided having to discuss the use of the assignment suppression operator, etc...

– David C. Rankin
Apr 14 '16 at 16:11

add a comment |

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