How to Print an array with names - c

Question

I wrote this code to input names to array (eg:Jon,raj...) and the input part is ok but how to print exact name that i input to this array a[5] why this code is not working
#include <stdlib.h>
#include<stdio.h>
int main(){
int i;
char a[5];
for(i=0;i<5;i++){
printf("Enter ");
scanf("%s",&a[i]);
}
for(i=0;i<5;i++){
printf("%s \n",a[i]);
}
}

Answer 1

Your array a is an array of chars. This means that in each position you will have only one char.
You can do this any one of these two ways:
Pre-allocated (array of arrays of chars)
#include <stdio.h>
int main(){
const int NAME_MAXIMUM_SIZE = 50;
const int NUMBER_OF_NAMES = 5;
int index;
char names [NUMBER_OF_NAMES][NAME_MAXIMUM_SIZE];
for(index=0; index<NUMBER_OF_NAMES; index++){
printf("Enter ");
scanf("%49s",names[index]);
}
for(index=0; index<NUMBER_OF_NAMES; index++){
printf("%s \n",names[index]);
}
}
Malloc + free (array of pointers)
#include <stdio.h>
#include <stdlib.h>
int main(){
const int NAME_MAXIMUM_SIZE = 50;
const int NUMBER_OF_NAMES = 5;
int index;
char *names [NUMBER_OF_NAMES];
for(index=0; index<NUMBER_OF_NAMES; index++){
names[index] = (char*)malloc(NAME_MAXIMUM_SIZE);
printf("Enter ");
scanf("%49s",names[index]);
}
for(index=0; index<NUMBER_OF_NAMES; index++){
printf("%s \n",names[index]);
free(names[index]);
}
}

Answer 2

You can make use of 2 dimensional array. Here your variable a which is a character array stores only one character at each index.
#include <stdlib.h>
#include<stdio.h>
int main(){
int i;
char a[5];
for(i=0;i<5;i++){
printf("Enter ");
scanf("%s",&a[i]);
}
for(i=0;i<5;i++){
printf("%s \n",a[i]);
}
}

Answer 3

"...wrote this code to input names to array..."
char a[5] does not create 5 C strings. It is simply an array of 5 char. It provides room for only 1, 4 character C string (leaving room for the \0 terminator). If you need an array of names, of typical size the code will need an array of arrays each with space sufficient for typical names. For example:
char a[5][80] = {{0}}; //provides an array capable of containing 5 names
Also, in the read statement: scanf("%s",&a[i]);, the format code %s is expecting to process an array of char. But is provided with &a[i] which is only a single character. If you are going to use scanf, then call it like this: scanf("%s",a); Same issue with the print statement.
Make the following edits to address these basic issues:
int main(void)
{
int i = 0;
char a[5][80] = {{0}};//space for 5 names, each with up to 79 characters
//arrays are initialized to zeros
for(i=0;i<sizeof(a)/sizeof(a[0]);i++)//sizeof(a)/sizeof(a[0]) is flexible way to
//loop only through number of names.
//If array size changes, expression will still work
{
printf("Enter name %d:\n ", i+1);
scanf("%79s",a[i]);//limit input to prevent overflow
// ^^
}
for(i=0;i<sizeof(a[0])/sizeof(a[0][0]);i++)
{
printf("%s \n",a[i]);
}
return 0;
}