Getting the first 10 characters of a string?

Question

I have not been able to find any information with a web-search. Where should I be looking?

@Ryan, Or leave it and mention that it's C++ since he might have meant that. — ikegami, May 02 '13 at 22:12
Avoid `strncpy`: https://randomascii.wordpress.com/2013/04/03/stop-using-strncpy-already/ — chqrlie, Dec 04 '16 at 13:11

score 26 · Accepted Answer · edited Jan 16 '18 at 14:05

26

char myString[256]; // Input string
char dest[256];     // Destination string

strncpy(dest, myString, 10);
dest[10] = 0; // null terminate destination

edited Jan 16 '18 at 14:05

Robb1

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answered May 02 '13 at 22:08

Lefteris E

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quick question, could I do strncpy(myString, myString,10) ? – user2341069 May 02 '13 at 22:14
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if you want to truncate the string to 10 chars then put myString[10] = 0; (This given that myString points to memory and not a string lettral) – Lefteris E May 02 '13 at 22:14
@user2341069 no you cant, as per the man page "The strings may not overlap, and the destination string dest must be large enough to receive the copy." As myString is a pointer to a char array in memory, using the same pointer twice means that they "overlap" in memory, and thus will cause an error. Essentially, this means that the src can not be a substring of dst or vice versa. – TonyArra May 02 '13 at 22:18
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Is the null terminate character not '\0' rather than just a 0? – TafT Feb 12 '16 at 11:24
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@TafT they are equivalent – Lefteris E Apr 29 '16 at 08:28
@LefterisE Thank you. It seems they are equivalent but '\0' is often used to indicate a character instead of a number. This question http://stackoverflow.com/questions/16955936/string-termination-char-c-0-vs-char-c-0 has a few sources in it. – TafT May 03 '16 at 14:45
I'm afraid I disagree: `strncpy()` has widely misunderstood semantics and is **very** error prone. **DO NOT USE THIS FUNCTION**. Learn why here: https://randomascii.wordpress.com/2013/04/03/stop-using-strncpy-already/ – chqrlie Dec 04 '16 at 13:00

score 7 · Answer 2 · answered May 02 '13 at 22:09

7

char source[] = "abcdefthijklmn";
char target[100];

strncpy(target, source, 10);
target[10] = '\0'; // IMPORTANT!

answered May 02 '13 at 22:09

tianz

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@michaelb958 So what do I need to add or change? – tianz May 02 '13 at 22:13
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target is not uninitialized pointer. target points to 100 bytes of memory in the stack. the \0 is not nessesary as strncpy will put it – Lefteris E May 02 '13 at 22:13
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@LefterisE Not sure about the initialization thing, but `strncpy` will not put a `'\0'` automatically – tianz May 02 '13 at 22:14
@LefterisE When I commented, the declaration was `char *target;`. – michaelb958--GoFundMonica May 02 '13 at 22:15
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@michaelb958 Yes it was. Thank you. I'm a bit confused with `char *a` and `char a[100]` – tianz May 02 '13 at 22:17
i am confused aswell about how source is declared. maybe source should have been char* as it is a pointer to a string literal – Lefteris E May 02 '13 at 22:18
@LefterisE I'm fairly sure that `char foo[] = "bar";` initialises `foo` with a copy of `"bar"`. – michaelb958--GoFundMonica May 02 '13 at 22:21
I'm afraid I disagree: `strncpy()` has widely misunderstood semantics and is **very** error prone. **DO NOT USE THIS FUNCTION**. Learn why here: https://randomascii.wordpress.com/2013/04/03/stop-using-strncpy-already/ – chqrlie Dec 04 '16 at 13:00

score 4 · Answer 3 · answered Dec 04 '16 at 13:10

4

Adding to the above answers:

char* someString = "your string goes here";

int main() 
{
  int n = 10;
  printf("(%.*s)\n", n, someString);

  return 0;
}

answered Dec 04 '16 at 13:10

StackUseR

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chqrlie · Answer 4 · 2016-12-04T13:08:05.227

There are many different ways to achieve your goal:

You can use snprintf (safest):

char source[] = "abcdefthijklmn";
char target[100];

snprintf(target, sizeof target, "%.10s", source);

You can use strncat if you know the destination has at least 11 elements:

char source[] = "abcdefthijklmn";
char target[100];

*target = '\0';
strncat(target, source, 10);

You can use strlen and memcpy (same assumption about the size of destination):

char source[] = "abcdefthijklmn";
char target[100];
size_t len = strlen(source);

if (len > 10)
    len = 10;
memcpy(target, source, len);
target[len] = '\0';

You can use a loop (same assumption about the size of destination):

char source[] = "abcdefthijklmn";
char target[100];
size_t i;
for (i = 0; i < 10; i++) {
    target[i] = source[i];
}
target[i] = '\0';

You could but should not use strncpy()

score 0 · Answer 5 · answered May 02 '13 at 22:12

If you're looking for a good source, here's an example of an online man page you can use: http://linux.die.net/man/3/strncpy

Important to note: you could also use memcpy instead of strncpy, but it requires you to add your own null-terminating byte.

Warning: If there is no null byte among the first n bytes of src, the string placed in dest will not be null-terminated.

Hence, memcpy and strncpy work almost the same here, and memcpy is more efficient and less prone to error.

Radar Blue · Answer 6 · 2016-08-28T18:30:31.583

I got it to work like this.

   # include <stdio.h>
   # include <string.h>
   //Strings. String lenght terminator.
   //KHO2016.no1. mingw (TDM-GCC-32) . c-ansi .
   int main ()
   {
   //declare
   char src_str[20],dst_str[10];

   //valuate
   printf ("Enter a sentance of 20 letters\n");
   gets (src_str);
   strcpy (dst_str,src_str);

   //calculate
   dst_str [10] ='\0'; // from the "www.stack overflow"
   printf ("%s",dst_str);
   printf ("\n");

   //terminate
   return 0;
   }

w.b · Answer 7 · 2016-12-04T13:00:49.373

0

You can also use sprintf with .10 precision format:

#include <stdio.h>
#include <string.h>

int main(void)
{   
    char source[] = "abcdefghijklmnopqrstuvwxyz";
    char dest[11];
    memset(dest, '\0', sizeof(dest));

    sprintf(dest, "%.10s", source);

    printf("%s", dest); // abcdefghij
    return 0;
}

edited Dec 04 '16 at 13:00

answered Dec 04 '16 at 12:00

w.b

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Getting the first 10 characters of a string?

7 Answers7