Printf prints more character than those contained in my string [closed] - c

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I have to write a program that acts like a shell. I wrote the function that gets the input from the user. I also wrote the function that splits it into arguments. The first time I type something, it works well, but the second time, it prints different characters after the ones that I gave it. I don't have to print it in the program. I was just doing it to see if it works correctly. I read a bunch of stuff online, but I can't figure out my error. I suppose it is in makeArgs(), but I can't pinpoint it.
Also, when I give it an input, the readline function adds a \n at the end of the string. I suppose it is from the fact that I press the enter key. I managed to solve the issue, by manually replacing it, but I would like to know if it is normal.
Any help really be appreciated.
Thank You
Screenshot of Xterm after 2 inputs.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int getText();
int makeArgs();
char *textEntre;
size_t nbCharacters;
char **arguments;
int main (void)
{
while (1){
getText();
int nbArguments = makeArgs();
for(int i =0; i<5; i++){
printf("%s \n",arguments[i]);
}
for(int i=0; i<nbArguments; i++){//free the char ptrs at the end
free(arguments[i]);
}
}
free(textEntre);
free(arguments);
return 0;
}
int getText(){
size_t buffersize = 0;
nbCharacters = getline(&textEntre, &buffersize, stdin);
textEntre[nbCharacters-1] =' '; // when I press enter it regiter the enter as \n so I replace it with a space
return 0;
}
int makeArgs(){
arguments = (char **)malloc(sizeof(char*)*20);
int i;
int j = 0;
int k = 0;
int nbElem = 20; //the number of ptrs that can be in arguments
for(i = 0; i<nbCharacters; i++){
if(i == 20){ //increases the memory allocated if there are more than 20 arguments
nbElem = nbElem *2;
arguments = (char **)realloc(arguments, sizeof(char*)*nbElem);
}
if(textEntre[i] == '"'){ //checks for ""
i++;
while(textEntre[i] != '"'){
i++;
}
}
if(textEntre[i] == ' ' && textEntre[i-1] == ' '){ // eliminates useless spaces
j++;
}
else if(textEntre[i] == ' '){ //save a single argument
char * chptr;
chptr = (char *)malloc(i-j+1); //giving +1 for the \0 at the end
strncpy(chptr, &textEntre[j], i-j);
arguments[k] = chptr;
k++;
j = i +1;
}
}
return k;
}

Answer 1

chptr = (char *)malloc(i-j+1); //giving +1 for the \0 at the end
You properly allocated memory for that terminating \0, but where do you actually add that "\0 at the end"?
strncpy(chptr, &textEntre[j], i-j);
strncpy does not necessarily zero-terminate the destination buffer. You have to do it yourself.
In fact, in this specific application strncpy is a rather inappropriate function: it does not give you anything over ordinary memcpy and might be less efficient. You could just do
memcpy(chptr, &textEntre[j], i - j);
with potentially better efficiency. And, again, don't forget to zero-terminate the destination buffer.
Or you can use sprintf for the same purpose as follows
sprintf(chptr, "%.*s", i - j, &textEntre[j]);
which will produce a properly zero-terminated string in the destination. (Albeit you won't see sprintf used that way very often.)