C : Copy shorter string into longer without using built in functions

Question

the task in this is to copy the first n characters from (null terminated) string2 (s2) into s1 while using pointers. I know how to copy the strings from one to another, but I am having issues with eliminating the extra values. For example:

s1= 'This is a test'
s2 = 'A test'

after copying, I am left with:

s1 = 'a tests a test'

Here is my code:

char *s1pointer;
const char *s2pointer;
int i;
int number_char_replace;

s1pointer = s1;
s2pointer = s2;
i=0;
number_char_replace = num; 

for(i=0;s1pointer[i] !='\0'||s2pointer[i]!='\0';i++)
{
s1pointer[i]=s2pointer[i];
}
}

This is homework, so please dont feel the need to just give me the answer. A hint for the logic would be greatly appreciated. Thank you.

I am surprised when OP says its homework and no one jumps out — SwiftMango
– SwiftMango, Commented Sep 2, 2013 at 19:19
I would say that the real problem here is that the OP didn't know that C strings need to be null-terminated. Now is that his fault, or the professor's? — Jonathon Reinhart
– Jonathon Reinhart, Commented Sep 2, 2013 at 20:52

cnicutar · Accepted Answer · 2013-09-02 18:56:47Z

6

You need to somehow indicate the end of the string after the for. Spoiler:

s1pointer[i] = 0;

answered Sep 2, 2013 at 18:56

cnicutar

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3 Comments

Jonathon Reinhart Over a year ago

Woah. I did not know you could do that!

Liquidmetal Over a year ago

Thank you for the help, I now have this section of my code outputting what I need.

Jonathon Reinhart Over a year ago

And on mobile, it says "Click for spoiler" (or something like that)

nickie · Accepted Answer · 2013-09-02 19:03:24Z

2

Three problems:

Your target string is not zero terminated when your code finishes. This can cause problems when, e.g., you try to print your string. Also, when you are copying a smaller string on top of a bigger string, it will be impossible to know where the new contents end and the older ones continue.
I understand why you're testing s2pointer[i] !='\0' in your for loop; you want to stop when the source string is exhausted. But why is s1pointer[i] !='\0' there too? Do you need it? Do you know anything about what s1 contains before the copying starts?
You said that you wanted to copy up to n characters. Where is that, in your code?

answered Sep 2, 2013 at 19:03

nickie

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1 Comment

Liquidmetal Over a year ago

The s1pointer was an oversight, it didn't hurt my previous outputs in functions so I included it. I forgot including N in the process of trying to figure it all out...I lost the forest in the trees I guess. Let me try to figure out the logic keeping the n in mind.

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