C programming: how do I store names in one dimensional array ?

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Example How many names: 3
Name 1: Jack
Name 2: Tom
Name 3: John
Those 3 names are: Jack Tom John.

What I have tried:

#include<stdio.h>
#include<string.h>
int i,n;
char s[50],array[50];
main(){
printf("How many names : ");
scanf("%d",&n);
for (i=0;i<n;i++){
printf("Name %d : ",i+1);
gets(s);

after that i tried to assign array[i]=s , it doesn't work
and i tried alot more but it keep displaying null.

Posted 23-Dec-17 21:40pm

Member 13590330

Updated 29-Jun-22 8:03am

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Patrice T 24-Dec-17 5:22am

What do you mean by:
'C programming: how do I store names in one dimensional array ?'
What do you want to do with the 3 names and the 1d variable?
Do you want to build a large string or simulate a 2d array ?

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Richard MacCutchan · Answer 1 · 2017-12-23T22:04:00

You read a string into the array s but you never do anything with it. If you want to save the names then you need to copy each one to its own array. the simplest way to do it is by creating a 2 dimensional array, something like:

C++

char names[4][16]; // 4 arrays each of 16 characters
// in the loop
for (i=0;i<n;i++)
{
    printf("Name %d : ",i+1);
    // gets(names[i]);
    fgets (names[i], 16, stdin);
}

However, since you do not know how many arrays there are in the first place you will need to do some dynamic allocation, and use character pointers. Something like:

C++

char** names; // a pointer to an unknown number of arrays
names = (char**)malloc(n * sizeof(char*)); // allocate the array pointers
// in the loop
for (i=0;i<n;i++)
{
    printf("Name %d : ",i+1);
    names[i] = (char*)malloc(16); // allocate an array for the next name
    //gets(names[i]);
    fgets (names[i], 16, stdin);
}

OriginalGriff · Answer 2 · 2017-12-23T21:57:00

The problem is that "names" are strings: each one is an array of char

char name1[] = "Jack";
printf("%s\n", name1);
char name2[] = "Tom";
printf("%s\n", name2);

So you can't easily keep them in a one dimensional array like yours.
What you need is a one dimensional array of pointer to char values, each of which is a separate name:

char* names[] = {"Jack", "Tom", "John"};
for (int i = 0; i < 3; i++)
   {
    printf("%s\n", names[i]);
   }

To read them in from the user is more complicated: you need to use malloc to allocate space for the data the user is going to type before each call to scanf, and store the pointer it returns in your names array.

merano99 · Answer 3 · 2022-06-29T08:03:00

Solution 5

C

int n = 0;
printf("How many names : ");
scanf("%d", &n);

Name* namelst = (Name*)malloc(n * NAMESIZE));

clearline();
    
for (int i = 0; i < n; i++) {
    printf("Name %d : ", i + 1);
    fgets(namelst[i], NAMESIZE , stdin);
}

with

C

#define NAMESIZE 25
typedef  char Name[NAMESIZE];
void clearline() 
{ for (int ch = 0; (ch != EOF) && (ch != '\n'); ch = fgetc(stdin)); }

Posted 29-Jun-22 8:03am

merano99

Comments

Patrice T 29-Jun-22 14:13pm

2017 !

merano99 29-Jun-22 17:17pm

No, it could have been coded like that in the early 80s. I gave Richard's dynamic solution a 5, but I also wanted to offer a simple alternative here with only one malloc.

The Task was to store names in one dimensional array. See Title.

C programming: how do I store names in one dimensional array ?

3 solutions

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