Print the string using dynamically memory allocated - c

Question

Hello, my question is: How to print out string 1 and string 2, not only string 2. I am new to dynamically memory allocated. My sample code is below, thanks for your help.
Result expected:
Hello
My name is Ken
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int main(){
int n, i;
char *ptr;
printf("How many strings you want to display?: ");
scanf("%d", &n);
ptr = (char*)malloc(n * sizeof(n));
if(ptr == NULL){
printf("Failed to allocate the memory to string!\n");
exit(0);
}
for(i = 0; i < n; i++){
printf("String %d: ", i + 1);
fflush(stdin);
scanf("%[^\n]", ptr);
}
printf("\n");
printf("Strings you entered:\n");
for(i = 0; i < n; i++){
printf("%s\n", ptr);
}
}

Answer 1

Please check the below code. I have marked modified lines with a comment starting with // CHANGE.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
// CHANGE - free memory when not needed
void free_memory(char** ptr, int n)
{
if (ptr)
{
int i;
for (i = 0; i < n; i++)
{
if (ptr[i])
{
free(ptr[i]);
}
}
free(ptr);
}
}
int main(){
int n, i;
char **ptr; // CHANGE - make a double pointer - array of strings (imagine rows and columns)
printf("How many strings you want to display?: ");
scanf("%d", &n);
ptr = (char**) malloc(n * sizeof(char*)); // CHANGE - allocate memory based on no. of rows (columns are string characters)
if (ptr == NULL) {
printf("Failed to allocate the memory to string!\n");
exit(1);
}
getchar(); // CHANGE - read the newline character else fgets doesn't work
for (i = 0; i < n; i++) {
ptr[i] = (char*) malloc(256 * sizeof(char)); // CHANGE - allocate memory for each row
printf("String %d: ", i + 1);
if ( fgets(ptr[i], 256 * sizeof(char), stdin) == NULL ) {
printf("Failed to get line input\n");
free_memory(ptr, n);
exit(1);
}
ptr[i][strlen(ptr[i]) - 1] = '\0'; // CHANGE - remove extra newline character at the end
}
printf("\n");
printf("Strings you entered:\n");
for (i = 0; i < n; i++) {
printf("%s\n", ptr[i]); // CHANGE - print each row
}
free_memory(ptr, n);
return 0;
}

Answer 2

First, there are two numbers that should be decided:
the number of strings the user wants to input and
the length of the strings (or the lengths of each string).
Each string will be stored in a char*. And all strings together will be stored in a char**. The length of char** will be your n in this case.
And length of each char* can be decided however you want.
I prefer to use a directive #define to define a constant like 100.
You can also do a global or local variable or just a digit.
Also don't forget to free anything you allocate. Check below for details.
And happy coding!
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAX_STRING_SIZE 100
int main(){
int n, i;
// stores all the strings
char **ptr;
printf("How many strings you want to display?: ");
scanf("%d", &n);
// allocate memory of size (n * size of char*)
ptr = (char**)malloc(n * sizeof(char*));
if(ptr == NULL){
printf("Failed to allocate the memory to string!\n");
// for portability to use EXIT_FAILURE,
// which is defined in the standard library
exit(EXIT_FAILURE);
}
// for each pointer, allocate memory of size (MAX_STRING_SIZE * size of char)
for(i = 0; i < n; i++)
// I should put a validity check here but I'm too lazy...
ptr[i] = (char*)malloc(MAX_STRING_SIZE * sizeof(char));
for(i = 0; i < n; i++){
printf("String %d: ", i + 1);
fflush(stdin);
scanf("%[^\n]", ptr[i]); // use ptr[i]
}
printf("\n");
printf("Strings you entered:\n");
for(i = 0; i < n; i++){
printf("%s\n", ptr[i]); // use ptr[i]
}
// free all of them
for(i = 0; i < n; i++)
free(ptr[i]);
free(ptr);
return 0;
}