While inputting the strings I am getting warnings like
error: format '%s' expects argument of type 'char *', but argument 2 has type 'char (*)[10]'
In application
#include<stdio.h>
int main(void)
{
char a[100][10];
int s,i,k=0;
scanf("%d",&s);
for(i=0;i<s;i++)
{
scanf("%s",&a[i]);
}
}
scanf("%s",&a[i]);
^ This will pass address of `a[i]` thus giving error.
As %s specifier expects a pointer to the first character in a character array. And using & with a[i] will evaluate char (*)[10].
Instead this should be done -
scanf("%9s", a[i]);
%s no problem in that.extra space because %9s will store just first 9 characters in array not more than that. So using %s will not harm you anyway .Yes because there is an difference between array and pointer to an array. You may use char *a[100]; as a pointer to array of 100 elements and use malloc/calloc every time to allocate space dynamically. The program would look like:
#include<stdio.h>
#include<stdlib.h>
int main(void)
{
char *a[100];
int s,i,k=0;
scanf("%d",&s);
for(i=0;i<s;i++)
{
a[i] = (char *) malloc (sizeof(char)*10);
scanf("%s",a[i]);
printf("%s",a[i]);
}
}
code: scanf("%s",&a[i]);
&a[i] point char **, because a is 2D char array.
code instead : scanf("%s",a[i]); will get right result
&a[i] is a char(*)[10], not a char**
%sviascanf, don't use the & with the variable name.