ansaurus

Question

Answer 1

A:

Homework?

Start with:

char a;
a = getchar();
printf("%c\n", a);

ysap 2010-08-12 01:38:45

He needs to learn... I wouldn't give him that answer if I were you.

thyrgle 2010-08-12 01:41:14

@thyrgle - well, at least it is not a complete solution...

ysap 2010-08-12 01:42:50

@ysap: I still wouldn't give him that much of hint without seeing a good try.

thyrgle 2010-08-12 01:45:10

@thyrgle: We're only under the assumption it's homework at the moment, although it seems like a good assumption we cannot be sure.

John K 2010-08-12 01:51:20

Answer 2

A:

#include <stdio.h>

int main()
{
    char c;

    printf("Input: ");
    scanf("%c", &c);

    printf("\n%i\n", (int)c);

    return 0;
}

Clark Gaebel 2010-08-12 01:41:27

Don't give him the full program! Can't you see this is his homework? >_<

thyrgle 2010-08-12 01:44:13

Relax. This is readily available all over the Internet.

Dumb Guy 2010-08-12 01:53:14

A downvote for answering the question? What's wrong with you people?

Clark Gaebel 2010-08-12 01:54:21

@Clark Gaebel: What downvote?

thyrgle 2010-08-12 01:58:51

Answer 3

A:

od -c > characters.txt

Charlie Martin 2010-08-12 01:59:16

Tagged as "c"..

Clark Gaebel 2010-08-12 02:09:24

and if comments had tags I would have tagged that "smart ass remark"

Charlie Martin 2010-08-12 02:11:01

Answer 4

+2 A:

Try this:

#include <stdio.h>


int main(int argc, char *argv) { 
    return argc == 1
        ? main(28, "the input given by the user") 
        : putch(argv[0]) && --argc>1 && main(argc, ++argv);
}

Anybody who turns this in as homework should probably work on the line: "Would you like fries with that?"

Jerry Coffin 2010-08-12 02:18:03

Shouldn't `argv` be a `char**`? In that case, the call to `main()` should be `main(6, "the input given by the user")` and you'd have to loop through an array of null terminated strings...

Chinmay Kanchi 2010-08-12 02:24:17

When the system calls it, it passes the address of an array of pointers to char -- but for that case, we only pay attention to `argc`. The implementation is not allowed to supply a prototype for main, so when we invoke it, we can pass what we feel like. If you think that's a bad idea, well, take note of the final sentence (and whoever downvoted, next time I'll try to remember to include closed-captioning for the humor impaired).

Jerry Coffin 2010-08-12 03:00:42

ansaurus

tags:

views:

answers:

How to find the ASCII Characters?

related questions