string not being read first time around - c

Question

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef char* string;
int main(void)
{
char *names[6];
int num_entries = 0,i=0,size=0;
string name = (string) malloc(sizeof(char) * 16);
printf("\nHow many names do you want to enter ? \n");
scanf("%d",&num_entries);
for(i=0 ; i < num_entries ; i++)
{
printf("\nEnter a name : ");
gets(name);
size = strlen(name);
names[i] = (string) malloc(sizeof(char)*size + 1);
strcpy(names[i],name);
}
for(i=0 ; i < num_entries ; i++)
puts(names[i]);
}
in this program the string is not read the first time around the loop,however works fine for all subsequent calls,the program simply has to accept n strings,store and display them. owever it executes n-1 times.Solution?also,feel free to point any mistakes in the way pointers,allocation etc. is used,any feedback appreciated .

Answer 1

Call gets before the loop to discard the new line left by scanf.
Or better yet, use the standard workaround to discard unread input:
int c;
while ((c = getchar()) != '\n' && c != EOF);

Answer 2

The issue here, which is typical of the scanf statement, is that it does not use the newline when you entered the number of names you wanted and pressed "enter".
As a result, the newline is stuck in the stdin buffer until you do your next read, which in this case is the first name you try to read, so your first name is simply "newline". To deal with this, use getchar() to eat up the newline character so you don't have that issue anymore.
Typically, as a rule of thumb, you'll almost always want to use a getchar() or something similar after a scanf statement to deal with this issue.
I've modified your code below and works fine for me. I also cleaned it up a bit since some lines were not necessary.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
typedef char* string;
int main(void)
{
string names[6];
int num_entries=0, i=0;
string name = malloc(sizeof(char) * 16);
printf("\nHow many names do you want to enter ? \n");
scanf("%d",&num_entries);
getchar();
for(i=0 ; i < num_entries ; i++)
{
printf("\nEnter a name : ");
fgets(name,16,stdin);
names[i] = malloc(sizeof(char)*strlen(name) + 1);
strcpy(names[i],name);
}
for(i=0 ; i < num_entries ; i++)
puts(names[i]);
return 0;
}

Answer 3

Here's the code with all suggestions. Note that Anthony Accioly gets credit for the answer.
int main(void)
{
char *names[6];
int num_entries = 0, i = 0, size = 0, c = 0;
string name = malloc(sizeof(char) * 16);
if ( !name )
{
printf( "Unable to allocate memory for name\n" );
return(1);
}
printf("\nHow many names do you want to enter ? \n");
scanf("%d",&num_entries);
while ((c = getchar()) != '\n' && c != EOF);
for( i = 0 ; i < num_entries; i++)
{
printf("\nEnter a name : ");
gets(name);
size = strlen(name);
names[i] = (string) malloc(sizeof(char)*size + 1);
strcpy(names[i],name);
}
for(i=0 ; i < num_entries ; i++)
puts(names[i]);
return(0);
}

Answer 4

You can also use fflush(stdin); as an alternative to getchar() or the while(...) statement.
P.S.: I am sorry for writing my suggestion here, as I do not have enough reputation to comment.