problem with strings (explained)

hi - i have a problem using strings in text output.

when i put the following code in my 'play'-function,

[...]

char name[20]; /* at top of the function */

printf("\n Hello! What is your name? > ");
scanf("%s",&name[0]);
printf("\n Ok, %s, let's get it on...\n\n",name);

[...]

then the program is running until you enter
a name like "Peter Pan" with a space inside.
Now the program would run once displaying
"Ok, Peter, ..." and after that (without
entering the 'again'-question) it runs with
"Ok, Pan, ...". and this way i don't want
to include this code in my program.

when i try this:

[...]

char name[20]; /* at top of the function */

printf("\n Hello! What is your name? > ");
gets(name);
printf("\n Ok, %s, let's get it on...\n\n",name);

[...]

the program is not working. it runs, but you won't
be able to enter your name. the program just displays
the text out of 'printf' and just puts a space instead
of the %s data.

can someone explain why this happens?

at last i tried this:

[...]

char name[20]; /* at top of the function */

printf("\n Hello! What is your name? > ");
gets(name);
printf("\n Ok, ");
puts(name);
printf(", let's get it on...\n\n");

[...]

but puts can't display anything, when gets won't work. :(

thanks for any help!

Code:

---

#include <stdio.h>

int main(void)
{
 char name[BUFSIZ];
 char surname[BUFSIZ];

 printf("Hello.  What is your name? \n> ");
 scanf("%s %s",&name,&surname);
 printf("Hello, %s %s, let's get it on...",name,surname);
 return 0;
}

---

:rolleyes: Is this one of those virtual sluts? :D

Your buffer still contains a '\n' at that point in the program it seems. Flush your buffer and it should work....

Code:

---

[...]
for (;getchar != '\n';);
gets(name);
[...]

---

FYI, the address-of operator (&) is not required when passing an array to scanf().

> for (;getchar != '\n';);

for ( ; getchar() != '\n'; ) ;
Make it a call...

Quote:

---

Originally posted by Deckard
FYI, the address-of operator (&) is not required when passing an array to scanf().

---

sorry, habit.

What if thier isn't 2 names? (In the surname exaple). Will it just ignore the second %s?

I had PRECISELY the same problem. Use gets () and all your problems will be solved!!! Also:

&n[0] == n

> Use gets () and all your problems will be solved
Certainly solves all the problems for anyone wanting to smash your program with a buffer overflow

Use fgets
http://www.eskimo.com/~scs/C-faq/q12.23.html

Yes, but this is not operating system code we are writing, here - although I see your point. The gets bug was the cause of the Unix worm - it's being discontinued.

Perhaps you'd better use fgets instead.