"Array" of strings

**Yorae** · 02-24-2006

I'm very new to C, and am studying pointers and memory allocation atm. It's fairly different from Java, so I'm a bit disoriented about it. Anyways, I have a problem that asks to read the number of students from the user and allocate memory locations that it equal to the number inserted. Here's the code I've done for that:

Code:

 
printf("enter the size of array:");
    scanf("%d",&s);

	names =(char*)malloc(s* sizeof(int));

Then I am asked to write a function that reads a name of a student and then use malloc to allocate exact space as the lenght to the string inserted.

Here's my code:

Code:

void addName(char*names)
{
    char *name;
	
	scanf("%s",name);
    
	char *copy = (char*) malloc(sizeof(char)*strlen(name));

	strcpy(copy,name);

	strcpy(names,copy);

	}

the program crashes on me everytime I run it. Also its supposed to return a pointer to the String, but I couldn't figure that out!

Any hint would be greatly appreciated!

**Ancient Dragon** · 02-24-2006

name is only an unallocated pointer so scanf("%s",name); will do no more than crash your program pretty hard. scanf() does not allocate the space, you have to do that yourself before calling that function

Code:

char name[255];
scanf("%s",name);

or more safely because it does not accept more characters
from the keyboard that can be put into the buffer.

fgets(name,sizeof(name),stdin);

The problem with fgets() is that it appends newline '\n' at the end

Code:

fgets(name,sizeof(name),stdin);
name[strlen(name)-1] = 0; // delete '\n'

in C it is not necessary to typecast the return value of any function that returns void*, like malloc(). And you forgot to add 1 more byte for the string's null terminator. No need for sizeof(char) because sizeof(char) is ALWAYS == 1.

Code:

char *copy =  malloc(strlen(name)+1);

Code:

strcpy(names,copy);

why ^^^^ ??? delete that line.

finally -- the function is supposed to return the pointer ? Then make the return type char*, not void

Code:

char* addName(void)
{

  ...
   return copy;
}

**Salem** · 02-24-2006

Well you could use a C compiler, rather than a C++ compiler "pretenting to be C"

char *name;
scanf("%s",name);
This fails for the same reason that you're trying to get past anyway, that a pointer doesn't point anywhere unless you malloc some space. So you initially just read your input to some random location.

char buff[100];
fgets( buff, sizeof buff, stdin );
Now do you strlen / strcpy etc

> char *copy = (char*) malloc(sizeof(char)*strlen(name));
Two things
1. C doesn't have mixed declarations and statements (C++ does)
2. There is no need to cast malloc in C, if you included stdlib.h

**eerok** · 02-24-2006

> C doesn't have mixed declarations and statements (C++ does)

C99 supports this; there may be good reason at times for not relying on its implementation on a given compiler, but it's C.

**fgw_three** · 02-24-2006

Originally Posted by eerok

> C doesn't have mixed declarations and statements (C++ does)

C99 supports this; there may be good reason at times for not relying on its implementation on a given compiler, but it's C.

I think gcc does a pretty good job with most of the C99 stuff. I waited a while until I was sure that what I wanted was supported, and then made the switch.

Of course, I don't have to worry too much about portability in my particular effort, so the usual "Your mileage may vary" will apply.

**dwks** · 02-25-2006

> char *copy = (char*) malloc(sizeof(char)*strlen(name));
Two things
1. C doesn't have mixed declarations and statements (C++ does)
2. There is no need to cast malloc in C, if you included stdlib.h

3. sizeof(char) is always 1, so you can leave it out

Remeber to free the memory when you're done with it.

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