Thread: Writing a code without arrays!

Hello, I am new to the C language. I am in need of help and/or advice on my code. I have to create a program that will transform a user's first and last name into their last name and their first name but with an initial. I am having trouble with writing the code and I will appreciate with advice. If this seems like I'm intending for someone to 'do my homework' then I apologize for the misunderstanding as that is not my intention.

Code:

#include <stdio.h>
int main();
{
    char ch;
    printf("Please enter your first and last name:");
    scanf("%d", &ch);
    printf("%d", &ch);
    getchar();
    return 0;
}

If you are trying to do strings without using arrays all I can say is...


Well, good luck with that.

Darn.... thanks!

A String is just an array of characters. You can either use an array like this

Code:

char ch[32];

or you can allocate memory to a pointer and treat it as an array.

So like Tater implied, what you want doesn't really make sense.

Anyways, your scanf and printf lines are already incorrect.

That depends what your definition of "array" is.

If it's just a variable that has brackets, your challenge is easy, just use malloc(), and it's sometimes not considered an array, just a space in memory that data is stored sequentially.

Otherwise, you might have to either do it character by character, or maybe mmap() using void..

Originally Posted by failure67

A String is just an array of characters. You can either use an array like this

Code:

char ch[32];

or you can allocate memory to a pointer and treat it as an array.

So like Tater implied, what you want doesn't really make sense.

Anyways, your scanf and printf lines are already incorrect.

I will put this to use and thank you for the pointers.

I shall provide some more information tomorrow. Thank you, memcpy and Tater for the help.

Originally Posted by CommonTater

You're kidding ... right?

About the only definition of "array" that even makes sense is "a sequential group of memory addresses used to store information".

No amount of semantic self-diddling is going to get around the simple fact that C stores strings in arrays...

This is technically true, but I also said "If it's just a variable that has brackets", meaning the "char d[3];" notation. I then said the "depending on the definition", because I've seen people (incorrectly) only consider bracket notation arrays, and malloc'd areas something else. I was trying to clarify whether the OP wanted to just avoid brackets and worded it wrong, or wanted to change the actual array part.

This is what I have so far. I'm not confident with how far I have gotten though :\

Code:

#include <stdio.h>
char main()
{
char ch1, ch2;
printf("Please enter your first and last name:");
scanf("%d", &ch1, &ch2);
printf("This is your name: %d", ch2, ch1);
 
getchar();
return 0;
}

You need to use an array of chars to hold a person's name. And you need to use %s as the scanf format specifier.

Code:

char main()

Incorrect, it should be

Code:

int main()

Maybe you should copy and paste your assignment. You may be misunderstanding it.

Except that it doesn't leave a space between the last name and initial. You might need to fix that.

Defining BLANK is worse than pointless. Just use ' '.