Question: In my code, what ever I enter for n, compiler allows me to input and output only half of it. Why?
#include<stdio.h>
#include<stdlib.h>
int main()
{
int n;
scanf("%d\n",&n);
char *c= (char*)malloc((n+1)*sizeof(char));
c[n]='\0';
for(int i=0;i<n;i++)
{
scanf("%c",&c[i]);
}
for(int i=0;i<n;i++)
{
printf("%c",c[i]);
}
}
Change this:
scanf("%c",&c[i]);
to this:
scanf(" %c",&c[i]);
Sample output:
Georgioss-MacBook-Pro:~ gsamaras$ gcc -Wall main.c
Georgioss-MacBook-Pro:~ gsamaras$ ./a.out
5
a
b
c
d
e
abcde
I discuss the reasoning behind this solution in Caution when reading char with scanf (C).
PS: Do I cast the result of malloc? No!
Also since you dynamically allocated memory, do not forget to free() it at the end of your main(), like this:
free(c);
Ques. compiler allows you to input only half of array size, Why?
Reason of problem :
scanf("%c",&c[i]) //Here %c takes enter key also as a part of input which reduces the input size to half.
So, there are mainly 2 solution to your problem:
Sol. 1 => you only need to include whitespace,rest code would be same.
scanf(" %c",c[i]) //use whitespace before %c
Sol. 2 => Don't enter one character at a time,please enter whole input as a bunch at once then press enter.
