1

it is not clear if + 1 is needed or not here:

int len = strlen(TARGET);
info = malloc( len + 1 );

because few lines above it it already was once appended to it:

TARGET[END - START] = '\0';

if it is needed then perhaps also.. appending the \0 is needed.

int len = strlen(TARGET);
info = malloc( len + 1 );
strcpy(info, TARGET);
info[len] = '\0';

Q: How to determine if a string already has the null termination

perhaps if it already has it.. appending another one wouldn't be logic.

full function :

char * FUNCTION ( char * v ){

char *TARGET = NULL;
const char *PATTERN1 = "co=";
const char *PATTERN2 = "&";
char *START = strstr(v, PATTERN1);

if (START) {
START = START + strlen(PATTERN1);
char *END = strstr(START, PATTERN2);
if (!END){
END = START + strlen(START);
}
TARGET = malloc(END - START + 1);
memcpy(TARGET, START, END - START);
TARGET[END - START] = '\0';
}

if (!START || TARGET == NULL || TARGET[0] == '\0') {
return 0;
}

int len = strlen(TARGET);
info = malloc( len + 1 );
strcpy(info, TARGET);
info[len] = '\0';

return info;
}
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8
  • 2
    "How to determine if a string already has the null termination" You couldn't use strlen if the string weren't null-terminated. Commented Jul 14, 2017 at 7:10
  • @FelixPalmen Hey, I'm not sure of other questions, but this looks like a legitimate question (I'm not judging the quality, though). Commented Jul 14, 2017 at 7:18
  • i am afraid there is a mistaken identity crisis in effect here perhaps. Commented Jul 14, 2017 at 7:22
  • my posts are being trolled by such invalid comments. Commented Jul 14, 2017 at 7:29
  • the damage is being done to the people that invested time to post answers. the damage is not done to the OP. the intent of the Poster is to solve a question. Commented Jul 14, 2017 at 7:31

3 Answers 3

Reset to default
5

How to determine if a string already has the null termination

Well, a "string", is by definition, null-terminated. Otherwise, it is not a string.

Quoting C11, chapter §7.1.1

A string is a contiguous sequence of characters terminated by and including the first null character. [....]

From theoretical point of view, it's the responsibility of the producer, not the consumer, to ensure the null-termination for a character array which is supposed to be used as string.

That said, strlen() returns the length of a string, without the null-terminator. So, if you were to use the return value of strlen() of an existing string to allocate memory for a copy thereof, you need to allocate one extra bye for the null-terminator, so the +1 is required while passing the size to allocator function.

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Comments

3

strlen calculates how many characters there are in a string up to and excluding the \0, so you'll never count it. So for example if you string is "hello", its strlen would be 5. So yes, you'll always need to add 1 to its length to account for the \0 at the end.

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2 Comments

if strlen returns a correct value, the string is null-terminated for sure...welcome to the world of UB.
Now that you say it, it seems obvious. I removed that part.
1

The following:

int len = strlen(TARGET);

will not give you sizeof(TARGET); it will just count the number of characters before the first \0. So, if TARGET contains "a\0bcd" it will give you just 1 (for 'a').

The following:

 info = malloc( len +1 );

Needs to allocate enough bytes for the string plus the null terminator, so the + 1 is necessary.

The following:

strcpy(info, TARGET);

will copy characters until the first '\0' is encountered, and then it will append a '\0', so there must be enough space in the destination for the '\0'.

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