Begginer with C, Help!

[fredsmith]

Hello,

 

So I am trying to make a program that simply asks the user for their name, then it greets them.

I am not sure what data type to assign to the variable.  Please see below.

#include <stdio.h>main(){????? name;printf("What is your name?");scanf("?????%", &name);printf("Welcome, ?????%", name);}

Appreciate your help!

 

Thanks.

[Builder]

char

[fredsmith]

char

I already tried that, it will only output the first character of the name, not the whole name.

[Borgdude]

If you're aware of arrays, trying using an array of chars.

[Builder]

I already tried that, it will only output the first character of the name, not the whole name.

Ah sorry. Forgot it's C and not C++. No string operators in C! You need a char array, like this:

unsigned char Name []

[Arphenyte]

You have to create an array. In this case, I believe it would be name...

 

char name[50];

 

And the scanf should be like this:

 

scanf("%s", name);

 

Same thing with the printf.

 

Notice there's no '&' in name since it's an array. If you don't know what an array is then I suggest going back to your teacher to ask how to make your program work since I don't see another path for this.

 

Hope it helps!

Edited September 7, 2014 by Lightz

[fredsmith]

 

Ah sorry. Forgot it's C and not C++. No string operators in C! You need a char array, like this:

unsigned char Name []

 

 

You have to create an array. In this case, I believe it would be name...

 

char name[50];

 

And the scanf should be like this:

 

scanf("%s", name);

 

Same thing with the printf.

 

Notice there's no '&' in name since it's an array. If you don't know what an array is then I suggest going back to your teacher to ask how to make your program work since I don't see another path for this.

 

Hope it helps!

 

 

Thanks! appreciate your help!

[Builder]

You have to create an array. In this case, I believe it would be name...

 

char name[50];

 

And the scanf should be like this:

 

scanf("%s", name);

 

Same thing with the printf.

 

Notice there's no '&' in name since it's an array. If you don't know what an array is then I suggest going back to your teacher to ask how to make your program work since I don't see another path for this.

 

Hope it helps!

Couldn't he just do

unsigned char []

and not have to worry about length?

[Arphenyte]

Couldn't he just do

unsigned char []

and not have to worry about length?

If he had used unsigned char [] then that means he would have to declare the name as well.

 

For example...

 

unsigned char name[]={"Joseph"};

 

My teacher always used to say you should ALWAYS declare array sizes UNLESS you declare it's content before hand, or unless you are working with pointers.

Edited September 7, 2014 by Lightz

[Builder]

-snip-

I've been spending too much time in scripting languages with type prediction. The information is all getting confuzzle'd together in my head. (oh Ruby, how I miss your ducks)

[reiderj8]

Couldn't he just do

unsigned char []

and not have to worry about length?

 

In C, you should always declare the array size. If you don't, you could run into some nasty seg faults errors that are not easy to debug. Another note for OP would be to always declare your char array to be a size of one larger than what you need because C always ends it's strings with a null terminator ('/0'). 

For Example if I declare a string as:

char string[] = {'t','e','s','t'}

when you try to print it, it will look something like this: testjhfdkhafhagfdhjkasd. It will print out garbage until it finds a null terminator. And this could also result in another seg fault error.

[Builder]

In C, you should always declare the array size. If you don't, you could run into some nasty seg faults errors that are not easy to debug. Another note for OP would be to always declare your char array to be a size of one larger than what you need because C always ends it's strings with a null terminator ('/0'). 

For Example if I declare a string as:

char string[] = {'t','e','s','t'}

when you try to print it, it will look something like this: testjhfdkhafhagfdhjkasd. It will print out garbage until it finds a null terminator. And this could also result in another seg fault error.

I should really just stop trying to write C...

 

I need to study it again I think. It's been too long.

[rashdanml]

My teacher always used to say you should ALWAYS declare array sizes UNLESS you declare it's content before hand, or unless you are working with pointers.

A fixed array length also has issues. If the size is discovered, using clever inputs, it can become a security issue (possibly fixable). 

[Ciccioo]

A fixed array length also has issues. If the size is discovered, using clever inputs, it can become a security issue (possibly fixable). 

that shouldn't be a "fixed array length" issue, but just a lack of control over the input

wether you use a fixed length array or a dinamically sized one, in the moment you call scanf() the array will be capable of storing a fixed amount of data, but the input could possibly go further

so the problem is not the data structure, but the method to access user input, and that's why scanf should be avoided for code that needs to be foolproof/safe (edit: actually there should be a format specifier to tell scanf the buffer length, but it's usually better to just use other functions)