Accept string using scanf and display them using printf

Question

I cannot retrieve string using printf . instead getting error : segmentation fault

int main(){

        char * a[5];
        int i;

        printf("\n enter value ");

        for (i=0;i<5;i++){
                printf("%d Name :\n",i);
                scanf("%s",&a[i]);
        }

        printf("%s",a[2]);
        printf("%s",a[3]);

}
~

You have an array of pointers and do not allocate any memore for them. Accessing memory via unitialized pointers is undefined behaviour and may result in seg fault. — Gerhardh, Apr 13 '18 at 10:42
Do you want to enter and print 5 names, or enter and print 1 name with 5 characters? Start with always allowing room for a NULL character in your string (char array), allocate it using a[6] instead of 5. — Grantly, Apr 13 '18 at 10:43

Achal · Answer 1 · 2018-04-13T12:18:38.907

char * a[5]; is array of 5 char pointer that means you can store 5 char buffer of any length. The problem is in the statement scanf("%s",&a[i]); replace this with scanf("%s",a[i]);

case 1 :-

int main(){
        char *a[5] = {"stackoverflow","meta","ask ubuntu","super","unix"};
        /* here a[row] can have any number of char */
        for (int i = 0 ;i < 5;i++) {
                printf("%s\n",a[i]);
        }
        return 0;
}

Case 2 :-

int main(){
        char *a[5];
        for (int i=0;i<5;i++){
            printf("%d Name :\n",i);
            a[i] = malloc(MAX);/* define MAX how many char you want to store into that */
            scanf("%s",a[i]); /* & not required as a[i] itself address */
        }
        for (int i = 0 ;i < 5;i++) {
                printf("%s\n",a[i]);
        }
        /* don't forget to free once job is done */
        for (int i = 0 ;i < 5;i++) {
                free(a[i]);
        }
        return 0;
}

score 0 · Answer 2 · edited Apr 13 '18 at 11:38

0

The problem is you are using a pointer before it has been initialized. It does not yet point at valid memory. These leads to program crashes or other kinds of unexpected behavior, such as "segmentation faults".

So you should edit it to:

char* my_string = malloc(size+1);

The %c format specifier is used whenever we want to be specific that the variable that we are going to printf or scanf is of type char.

On the other hand the %s format specifier is used to specify to the printf or scanf functions that contents of the address that is specified as the next parameter are to considered as string.

You are taking char as input, not the whole string.

scanf("%s",&a[i]);, will be scanf("%c",&a[i]);
printf("%s",a[3]);, will be printf("%c",a[3]);
printf("%s",a[2]);, will be printf("%c",a[2]);

The following code worked fine:

#include<stdio.h>
#include<stdlib.h>

int main(){

        char* a = malloc(size+1); // where size is max string length
        int i;

        printf("\n enter value ");

        for (i=0;i<5;i++){
                printf("%d Name :\n",i);
                scanf("%c",&a[i]);
                printf("%c\n",a[i]);
        }

        printf("%c\n",a[2]);
        printf("%c\n",a[3]);
        free(a);
}

edited Apr 13 '18 at 11:38

Lundin

195,001
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answered Apr 13 '18 at 10:36

Abhishek Keshri

3,074
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"worked perfectly fine" is one of many results of undefined behavior. You really shouldn't teach beginners to store 1 char in one of the bytes of a pointer, then print it as if it was the correct thing to do. The OP obviously don't have a clue how to use pointers or strings in the first place, so this isn't helping. – Lundin Apr 13 '18 at 11:20
Yes, you are right, I will edit it as one of the methods and wite some basics of pointers. – Abhishek Keshri Apr 13 '18 at 11:22
Thanks for the suggestion – Abhishek Keshri Apr 13 '18 at 11:22
Hmm. You _have_ noticed that the OP used a `char*` array and not a `char` array, right? – Lundin Apr 13 '18 at 11:24
Yeah, I am editing the post regarding that only. – Abhishek Keshri Apr 13 '18 at 11:25
is it okay now? or still i am missing something? – Abhishek Keshri Apr 13 '18 at 11:29
I took the liberty to edit your post and add the malloc call you proposed to the snippet at the bottom. – Lundin Apr 13 '18 at 11:39
thanks for that, that way it's better – Abhishek Keshri Apr 13 '18 at 11:41

con ko · Answer 3 · 2018-04-13T11:15:26.433

0

You shouldn't input to a char* . And you shouldn't store any string by char*, including string literal, in fact, string literal convert to char* is deprecated many years ago. Use char a[5][MaximumLength] instead. MaximumLength should be a literal.

edited Apr 13 '18 at 11:15

answered Apr 13 '18 at 10:42

con ko

1,368
5
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Accept string using scanf and display them using printf

3 Answers3