Printing the contents of uneven (non-quadric) matrix in C - c

Question

The goal of my assignment (ONLY C, NO C#,CPP ect...) was to get strings from the user, then trim the matrix to fit the exact data. This is my code:
void main(){
char** text=Q1();
int i;
printf("Lines entered:\n");
for(i=0;i<sizeof(text)/sizeof(char*);i++)
printf("%s\n",text[i]);
}
char** Q1(){
char OOM[]="Out of memory! exiting.";
char **buf=(char**)calloc(BUFFER,sizeof(char*)),**text;
int i,j;
if(buf==NULL){
printf(OOM);
exit(0);
}
for(i=0;i<BUFFER;i++){
buf[i]=(char*)calloc(BUFFER,sizeof(char));
if(buf[i]==NULL){
printf(OOM);
exit(0);
}
}
i=0;
do{
i++;
printf("Enter string #%d:\n",i);
gets(buf[i-1]);
}while(strlen(buf[i-1])>0 && i-1<BUFFER);
text=(char**)calloc(i,sizeof(char*));
if(text==NULL){
printf(OOM);
exit(0);
}
for(j=0;j<i;j++){
text[j]=(char*)calloc(strlen(buf[j]),sizeof(char));
if(text[j]==NULL){
printf(OOM);
exit(0);
}
strcpy(text[j],buf[j]);
}
free(buf);
return text;
}
Now I know this could be done with realloc (which I am not quite familiar with just yet).
I want to test that my allocations are okay and print the strings I saved in text.
What is the most intuitive way of doing that?
Thanks in advance!

Answer 1

First, this code:
for(i=0;i<sizeof(text)/sizeof(char*);i++)
printf("%s\n",text[i]);
Will not work.
sizeof(text) will return the size of a pointer since text is a char**. You'd want something like:
for (int i=0; i<BUFFER && text[i]!=0; i++)
printf("%s\n",text[i]);
But this requires that the last entry in text is a null pointer so you would need to allocate 1 more entry than required and set it to NULL.
Otherwise, you'll need to find some other way to return the size.
Next:
i=0;
do{
i++;
printf("Enter string #%d:\n",i);
gets(buf[i-1]);
}while(strlen(buf[i-1])>0 && i-1<BUFFER);
i will be 1 more than the number of entries. For example, if the user presses ENTER on the first try, i will be set to 1.
Why not?
for (i=0; i<BUFFER; i++) {
printf("Enter string #%d:\n",i+1);
gets(buf[i]);
if (strlen(buf[i])­==0) {
break;
}
}
This way i is set to the number of entries.
Assuming you make the change above then:
text=(char**)calloc(i,sizeof(char*));
Will need to be changed to:
text=(char**)calloc(i+1,sizeof(char*)); // Include an extra entry for the sentinel
text[i] = 0; // Add sentinel to indicate last entry
Also, when you allocate the space for the strings, you need to add 1 to the size to leave room for the terminating null character. So
text[j]=(char*)calloc(strlen(buf[j]),sizeof(char));
becomes
text[j]=(char*)calloc(strlen(buf[j])+1,sizeof(char));
And you forgot to free buf[0] to buf[BUFFER-1]
For the real question. I'm not sure what's expected your approach looks OK to me.