c - 如何将偶数个字符复制到C中的字符串？

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" .

Answer 1

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.

Answer 2

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);