Can't dynamically allocate a string - c

Question

I tried to dynamically allocate a string using a function I named ALLO, but when I execute I get an error, which is my function ALLO can't get the string using getc, it gets skipped.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void ALLO(char *str){
char c=0;
int i = 0, j = 1;
str = (char*)malloc(sizeof(char));
printf("Enter String : ");
while (c != '\n') {
// read the input from keyboard standard input
c = getc(stdin);
// re-allocate (resize) memory for character read to be stored
str = (char*)realloc(str, j * sizeof(char));
// store read character by making pointer point to c
str[i] = c;
i++;
j++;
}
str[i] = '\0'; // at the end append null character to mark end of string
printf("\nThe entered string is : %s", str);
free(str); // important step the pointer declared must be made free
}
int main(){
char *NomAF;
int NAF;
printf("Entrer le nombre des ateliers : ");
scanf("%d",&NAF);
ALLO(NomAF);
return 0 ;
}

Answer 1

The semantics are wrong.
You ask the user for the names of the athletes, and then you scan it into an integer. You should ask for the number of athletes first. Then, after that, you allocate memory to accommodate each name.
int num_names;
scanf("%d", &num_names);
After you know the number of names, you then allocate a buffer for each name, separately.
char **names;
names = malloc(num_names * sizeof(char **));
for(int i = 0; i < num_names; i++)
ALLOC(&names[i]);
Also, you shouldn't be using scanf for user input. Use fgets instead, which is a little better.
Then, you also should be using a pointer to pointers to get those strings.
A little modified version of your code (which you should review and fix, as needed):
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void ALLO(char **str){
/* use INT for getc() return */
int c=0, i = 0;
/* you are gettting 1 byte of memory */
*str = malloc(sizeof **str);
/* should use fprintf(stderr...) or fflush(stdout) to guarantee
* the sentence will be seen by user
*/
printf("Enter String : ");
while (c != '\n') {
// read the input from keyboard standard input
c = getc(stdin);
// re-allocate (resize) memory for character read to be stored
/* i = 0 in the first run,
*
* and you have 1 byte alloced in the first run.
*
* so you get 1 byte for actual getc() return
* 1 byte for next character + NULL byte
*
* NOTE: you are STORING the NULL byte in your string. You only
* check for it AFTER you do the assignment, so your strings
* contain a newline before the NULL byte.
*/
*str = (char*)realloc(*str, (i + 2) * sizeof **str);
// store read character by making pointer point to c
(*str)[i] = c;
// you can use only 'i' for this...
i++;
/* #i
*
* Using only 'i' requires that you understand what #i is doing
* during execution. #i will keep the current buffer position,
* and you know you need one more position for the next
* character and one more for the NULL byte.
*
* Therefore, in your realloc statemente, you need #(i + 2)
*/
}
(*str)[i] = '\0'; // at the end append null character to mark end of string
printf("\nThe entered string is : %s", *str);
// if you free here, you can't get the string at #main for printing.
// free is the last step
//free(str); // important step the pointer declared must be made free
}
int main(){
char **NomAF;
int NAF, i;
char buf[100];
printf("Number of athlets : ");
fgets(buf, sizeof(buf), stdin);
NAF = atoi(buf);
NomAF = malloc(NAF * sizeof *NomAF);
// check malloc errors
// get names
for(i = 0; i < NAF; i++) {
ALLO(&NomAF[i]);
printf("New name: %s\n", NomAF[i]);
}
// print names, then free() then
for(i = 0; i < NAF; i++) {
printf("Name: %s\n", NomAF[i]);
free(NomAF[i]);
}
// free the base pointer
free(NomAF);
return 0 ;
}

Answer 2

Add this
while((c=getchar()!='\n')&&c!=EOF);
before getc it skips white space.
Because after this scanf("%d",&NAF); take you are giving input 5(+enter) this goes 5'\n' 5 is got by scanf and '\n' is in buffer and this new line is got by your getc.
and change this str[i] = '\0'; to str[i-1] = '\0';, it replaces the newline with NULL and you allocated memory for i characters only.
You can return the string by return str; and change function return type as char* or if don't want that take a parameter that allocated by malloc.
See this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void ALLO(char *str)
{
char c=0;
int i = 0;
printf("Enter String : ");
while (c != '\n')
{
while((c=getchar()!='\n')&&c!=EOF);
c = getc(stdin);
str = (char*)realloc(str, (i+1) * sizeof(char));
if(!str) exit(1);
str[i] = c;
i++;
}
str[i-1] = '\0';
}
int main()
{
char *NomAF;
int NAF;
NomAF=malloc(sizeof(char));
if(!NomAF) exit(1);
printf("Entrer le nombre des ateliers : ");
scanf("%d",&NAF);
ALLO(NomAF);
printf(NomAF);
free(NomAF);
return 0 ;
}
output:
Entrer le nombre des ateliers : 5
Enter String : a
s
d
f
g
----->newline to stop the loop
asdfg
Process returned 0 (0x0) execution time : 8.205 s
Press any key to continue.
I entered it as a string not character by character, its not practical to ask the user the enter letter by letter
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
char * inputword(char **);
int main()
{
char *word;
printf("Enter Word:");
word=inputword(NULL);
printf("Word Entered:%s",word);
free(word);
printf("\nEnter Word 2:");
inputword(&word);
printf("Word Entered:%s",word);
free(word);
return 0 ;
}
char *inputword(char **word)
{
char *str=NULL,ch,*memerr="Memory Error";
int i=0,flag=1;
do
{
str=realloc(str,((i+1)*sizeof(char)));
if(!str)
{
printf(memerr);
exit(1);
}
ch = getch();
switch(ch)
{
case 13:
str[i] = '\0';
putc('\n',stdout);
flag=0;
break;
case '\b':
if(i>0) i--;
str[i--]='\0';
printf("\b \b");
break;
default:
str[i] = ch;
putc(ch,stdout);
}
i++;
}while(flag);
if(word!=NULL)
*word=str;
return str;
}
output:
Enter Word:Hai, How are You?(1 String)
Word Entered:Hai, How are You?(1 String)
Enter Word 2:Hai, How are You?(2 string)
Word Entered:Hai, How are You?(2 string)
Process returned 0 (0x0) execution time : 58.883 s
Press any key to continue.