Kamikaze_Badger Posted August 4, 2005 MinGW is a C and C++ compiler. Get Dev-C++, it comes with MinGW and has a much better interface. Quote Share this post Link to post Share on other sites
Blue_cow Posted August 4, 2005 (edited) Anybody want to give me a challenge? to make a program to do something. BTW I AM A MAJOR NOOB! I STARTED LEARNING C TODAY! I KNEW NOT ONE SINGLE THING ABOUT IT BEFORE TODAY. so give me a program to make. Edited August 4, 2005 by Blue_cow Quote Share this post Link to post Share on other sites
Kamikaze_Badger Posted August 4, 2005 (edited) Have the user enter a sentence and then redisplay it. That'll test your usage of variables, arrays, and the printf() and scanf() functions. Edited August 4, 2005 by Kamikaze_Badger Quote Share this post Link to post Share on other sites
Blue_cow Posted August 4, 2005 (edited) Heres what ive got so far but im kindof stuck... #include <stdio.h> int main () { char sentance; printf ("Please enter a sentance:"); scanf ( "%s", &sentance ); printf ("You typed: %c\n", sentance); return 0; } EDIT: Ok, i got it working past debug mode but when it asks to enter a sentance, you do, and when you press enter it closes... :/ Edited August 4, 2005 by Blue_cow Quote Share this post Link to post Share on other sites
markiemrboo Posted August 4, 2005 Heres what ive got so far but im kindof stuck... #include <stdio.h> int main () { char sentance; printf ("Please enter a sentance:"); scanf ( "%s", &sentance ); printf ("You typed: %c\n", sentance); return 0; } EDIT: Ok, i got it working past debug mode but when it asks to enter a sentance, you do, and when you press enter it closes... :/ 522317[/snapback] Try system("pause"); at the end Like... printf ("You typed: %c\n", sentance); system("pause"); return 0; } I think you may have to include "stdlib.h" too, so up where it says #include <stdio.h> add another line: #include <stdlib.h> Quote Share this post Link to post Share on other sites
Kamikaze_Badger Posted August 4, 2005 (edited) Also, try using a char array if you want more then one letter. Or, if you're lazy: typedef char[] string Edited August 4, 2005 by Kamikaze_Badger Quote Share this post Link to post Share on other sites
markiemrboo Posted August 4, 2005 Also, try using a char array if you want more then one letter. Or, if you're lazy: typedef char[] string 522404[/snapback] Well picked up on! Buuut your solution doesn't solve the problem, nor does it actually work char sentance[500]; Should suffice for something basic, just so long as you don't go over 500 characters on your input! Quote Share this post Link to post Share on other sites
Kamikaze_Badger Posted August 4, 2005 What did I do wrong now? Quote Share this post Link to post Share on other sites
Blue_cow Posted August 4, 2005 (edited) You two are like a married couple. Thanks for the help guys! Ill see if it works. Edit: here's what i have, and it doesnt work. #include <stdio.h> #include <stdlib.h> int main () { char sentance[500]; printf ("Please enter a sentance:"); scanf ("%s", &sentance ); printf ("You typed: %c\n", sentance); system("pause"); return 0; } It compiles with no bugs, but when you enter in a senctence (eg. This is a senctence) it will come back with: You typed: p. Every time. So... yeeeahh. Edited August 4, 2005 by Blue_cow Quote Share this post Link to post Share on other sites
markiemrboo Posted August 4, 2005 You two are like a married couple. Thanks for the help guys! Ill see if it works. Edit: here's what i have, and it doesnt work. #include <stdio.h> #include <stdlib.h> int main () { char sentance[500]; printf ("Please enter a sentance:"); scanf ("%s", &sentance ); printf ("You typed: %c\n", sentance); system("pause"); return 0; } It compiles with no bugs, but when you enter in a senctence (eg. This is a senctence) it will come back with: You typed: p. Every time. So... yeeeahh. 522447[/snapback] eek, missing a lot today! Simple error... printf ("You typed: %c\n", sentance); Change %c (which prints out only one character) tooo %s. printf ("You typed: %s\n", sentance); aaaaand KB: Did you try and use that typedef yourself? AFAIK [] is just a friendly was of saying it's a pointer. I could be wrong there so feel free to look it up! (I am kinda busy). Buuut arrays have to be after a variable name it seems, so using it in a typedef wouldn't work..? Even if it did work - assuming it is just a friendly way of saying "I'm a pointer" - instead of having storage space for one char you've got a pointer to some random place in memory, and the sentance would be stored.... *somewhere*. It *may* well work and not crash, but you'll have unpredictable results by overwriting random memory I think Quote Share this post Link to post Share on other sites
Kamikaze_Badger Posted August 4, 2005 (edited) Lol, I'm monogamous and have myself a significant other already. Markie can wait . EDIT: Markie, you posted at the same dang time as me. Lemme put some more in... EDIT2: dang, I was always told that char[] daArrayz0rz = "Da arrayz0rz!" would tell the compiler to fill in the blank itself... dang... Would this work? typdef char string [1024] Edited August 4, 2005 by Kamikaze_Badger Quote Share this post Link to post Share on other sites
Blue_cow Posted August 4, 2005 [quote name=markiemrboo' date='Aug 4 2005, 12:55 PM printf ("You typed: %c\n", sentance); Change %c (which prints out only one character) tooo %s. printf ("You typed: %s\n", sentance); 522452[/snapback] BAHH!!!! How did i miss that! Ok so (sorry if it feels like you guys are walking me through it, i think im maknig a bit of progress.) now it works except that the output is only one word. Do i need to have more scanfs and/or variables? Quote Share this post Link to post Share on other sites