Accept string using scanf and display them using printf [duplicate] - c

Question

This question already has answers here:
Crash or "segmentation fault" when data is copied/scanned/read to an uninitialized pointer
(5 answers)
Closed 4 years ago.
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]);
}
~

Answer 1

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;
}

Answer 2

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.

Answer 3

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);
}