How to copy an even number of char to a string in C?

Question

I'm trying to copy a even number of char in another string. It does not work at all. But if I try to copy a odd number of char it works. I try to use a for and the function strncpy and it print the word that I want plus a randome letter.

int position=6;
char *stringFirst=malloc(position);

for(int j=0;j<position;j++){
    stringFirst[j]=fileStingToMod[j];
}

printf("%s",stringFirst);
free(stringFirst);

This is another code that I try to run:

int position=6;
char *stringFirst=malloc(position);

strncpy(stringFirst,fileStingToMod,position);

printf("%s",stringFirst);
free(stringFirst);

In both cases, the code gives me the following output:

cammelÙ

or

cammel↓

The string, named fileStringToMod, is : "cammelloverde".

If you want to apply string-format operations on that array of characters, then you set the last character to 0 (null character). — goodvibration, Aug 31 '20 at 14:29
`printf("%s",stringFirst)` is an example of a string-format operation. — goodvibration, Aug 31 '20 at 14:30
Thanks a lot man, I'm really dumb. But I have a question: Why my code worked with odd nomber of character and not with even number of character? — , Aug 31 '20 at 14:37
@LudovicoLatini undefined behavior does not need to provide wrong results. That's the bad thing about it. It has nothing to do with the number being odd. — RobertS supports Monica Cellio, Aug 31 '20 at 14:38

RobertS supports Monica Cellio · Accepted Answer · 2020-08-31T14:59:47.400

3

Only the first six characters of fileStringtoMod are copied to the memory pointed by stringFirst.

Note that a null terminator is missing to determine the end of a string.

The %s format specifier is to print strings, which need to be determined by null, but stringFirst has no null terminator.

Using printf("%s",stringFirst); invokes undefined behavior.

Use

stringFirst[position-1] = '\0';

before

printf("%s",stringFirst);

if you only want to print "camme".

Alternatively, you could use

printf("%.*s", position, stringFirst);

as suggested by @IngoLeonhardt the comments or

printf("%.6s", stringFirst);

even without having a null terminator in the string.

If you want to print 6 characters (like "cammel") you need to allocate memory for 7, not 6 characters as you always need a null terminator to representing strings.

edited Aug 31 '20 at 14:59

answered Aug 31 '20 at 14:34

RobertS supports Monica Cellio

14,524
7
33
80

1

as an addition: `printf("%.*s", position, stringFirst);` would be ok even without `stringFirst` beiing nul-terminated – Ingo Leonhardt Aug 31 '20 at 14:48
You've got a typo now: it's`%.6s` – Ingo Leonhardt Aug 31 '20 at 14:52
@IngoLeonhardt Ah, I knew that this is possible but didn't thought about it. Thank you for hinting. ;-) – RobertS supports Monica Cellio Aug 31 '20 at 14:54
It *is* necessary. Without the point, `%6s` will add additional space if the string is shorter but it won't truncate so nul is still needed – Ingo Leonhardt Aug 31 '20 at 14:58
@IngoLeonhardt Need to do more investigation about this. I didn't used such string printing very often. Thank you again. Corrected. – RobertS supports Monica Cellio Aug 31 '20 at 15:00
@chux-ReinstateMonica OK. Thank you. I will take more care about the difference. – RobertS supports Monica Cellio Aug 31 '20 at 15:05

score 0 · Answer 2 · answered Aug 31 '20 at 14:39

0

You didn't specify the contents of FileStingToMod, but I just guess it is a longer string.

The problem is that in C strings are terminated by a '\0' character, which you are not adding to your new string.

int position = 6;
char *stringFirst = malloc(position);
for(int j=0; j<position-1; j++) {
    stringFirst[j]=fileStingToMod[j];
}
stringFirst[position-1] = '\0';
printf("%s",stringFirst);
free(stringFirst);

answered Aug 31 '20 at 14:39

Alejandro López

144
5

I solved the problem. It works with stringFirst[position] = '\0'; and not with stringFirst[position-1] = '\0'; – Aug 31 '20 at 14:42
4

@LudovicoLatini If you only use `stringFirst[position]` you again have undefined behavior because `stringFirst[position]` does not exist. Array indices start at `0`; not `1`. – RobertS supports Monica Cellio Aug 31 '20 at 14:43
@SystemCrash Regarding your first sentence: You can find the content at the end of the question post, which also gives explanation for what's going on here. – RobertS supports Monica Cellio Aug 31 '20 at 14:46
I know but I rely on what I see running the program. – Aug 31 '20 at 14:47
2

@LudovicoLatini That's the problem. Just because it shows right at one specific execution on your system, it doesn't mean your code is broken. That's what I meant in my comment to your question as well. – RobertS supports Monica Cellio Aug 31 '20 at 14:48
@LudovicoLatini Beside that: Is `stringFirst[position-1] = '\0';` not working? – RobertS supports Monica Cellio Aug 31 '20 at 14:48
@RobertSsupportsMonicaCellio Nope it does not work, it prints 5 letters of my string if position is 6. – Aug 31 '20 at 14:53
1

@LudovicoLatini It's correct that it prints only 5 characters, since you only allocated 6 characters. 1 character is needed for the null terminator, which is not printed. – RobertS supports Monica Cellio Aug 31 '20 at 14:57
So, if you want to display 6 characters, you need to set position=7 – Alejandro López Sep 01 '20 at 15:05

How to copy an even number of char to a string in C?

2 Answers2