Pointers and strncat

Hello,

I'm trying to keep the firstLine char pointer to be independent from the spaces one but i don't manage to achieve that.

Code:

char* firstLine = spaces;
strncat(firstLine,s,m-rm); //that's it now I want the actual block on memory to be copied.

			s = malloc(sizeof(int)*(n-m)); j++;
			for (k=0; k<n; k++){
				s[k] = A[j]; j++;	  
			} char* rString = s; A = spaces;
			strncat(A,rString,n-rm);

So by the end firstLine and spaces should point to different places in memory. How to achieve that?

Can you be more specific in explaining your problem, the one you gave is kind of vague.

Well, lets look at this one then:

Code:

char* breakSmart(char* A, int n, int m, int* costs){
	char* formatted = malloc(sizeof(char)); char* not = A; char* rFormat; char* tempNot; int j; int inLength, rLength;
	if (n > m)
		while ((inLength = length(not)) > m){
			j = inLength;
			while (not[j] != '\n' && not[j] != ' ') j--;
			not[j] = '\n';rLength = inLength-j;
			rFormat = malloc(sizeof(char)*rLength);  int r;
			for (r=j+1; r<=inLength; r++)
				rFormat[r] = not[r];
}

For some weir reason rFormat ends up with nothing, although not/A contains for them.

no one to help on this here?

Originally Posted by simpatico_qa

no one to help on this here?

It would help if you formatted your code better, e.g.,

Code:

char* breakSmart(char* A, int n, int m, int* costs) {
    char* formatted = malloc(sizeof(char));
    char* not = A;
    char* rFormat;
    char* tempNot;
    int j;
    int inLength, rLength;
    if (n > m)
        while ((inLength = length(not)) > m) {
            j = inLength;
            while (not[j] != '\n' && not[j] != ' ') j--;
            not[j] = '\n';
            rLength = inLength - j;
            rFormat = malloc(sizeof(char) * rLength);
            int r;
            for (r = j + 1; r <= inLength; r++)
                rFormat[r] = not[r];
        }

So, it turns out that your code is invalid.

Basically, you should post code that you attempt to compile, and also post what is the problem.

with me it compiles but simply doesn't fill the char array. What makes it invalid to you?

Originally Posted by simpatico_qa

What makes it invalid to you?

The function is a fragment. You do not have the closing brace at the end, and you are missing the return statement (though that might result in a warning rather than an error, but is a mistake nonetheless).

that's because i tried to post the code that is problematic. So if u close the brackets and put a return u'll still see that the rFormat is not filled. This is my problem.

that's because i tried to post the code that is problematic. So if u close the brackets and put a return u'll still see that the rFormat is not filled. This is my problem.

Okay, so you are saying that we can take this as the function implementation:

Code:

char* breakSmart(char* A, int n, int m, int* costs) {
    char* formatted = malloc(sizeof(char));
    char* not = A;
    char* rFormat;
    char* tempNot;
    int j;
    int inLength, rLength;
    if (n > m)
        while ((inLength = length(not)) > m) {
            j = inLength;
            while (not[j] != '\n' && not[j] != ' ') j--;
            not[j] = '\n';
            rLength = inLength - j;
            rFormat = malloc(sizeof(char) * rLength);
            int r;
            for (r = j + 1; r <= inLength; r++)
                rFormat[r] = not[r];
        }
    return rFormat;
}

Now, what is your test input? What is the expected test output and actual test output?

Remember, if n <= m, then rFormat will be left uninitialised. Also, by assigning to rFormat in a loop, it means that rFormat will only point to the memory allocated on the last iteration of the loop (and the memory allocated on all previous iterations will be leaked).

There is something clearly absent from my mind:

Code:

int* init(){
	int* costs = malloc(sizeof(int)*co);
	int i=0;
	while (*(costs+i) != '\0'){
		costs[i] = INF;
		i++;
	} return costs;
}

Even this doesn't not enter the while loop. What's up?

The array is uninitialised, so you should not be accessing it like that (especially since the point of the loop is to initialise the array, is it not?)

is not it initialized by default to 0? Would u be so kind to articulate ur suggestion in code so that i grasp it?

Originally Posted by simpatico_qa

is not it initialized by default to 0? Would u be so kind to articulate ur suggestion in code so that i grasp it?

No, it is not. You are probably thinking of calloc() instead of malloc(). But if you use calloc(), the loop will terminate at the start of the first iteration.

What are you trying to do?

just fill the costs array with a fixed no, say 34.

Originally Posted by simpatico_qa

just fill the costs array with a fixed no, say 34.

Right. Take a look at this example:

Code:

#include <stdio.h>
#include <stdlib.h>
#include <stddef.h>

int *initCosts(size_t size, int initial_value);

int main(void)
{
    const size_t size = 10;
    const int initial_value = 34;
    int *costs = initCosts(size, initial_value);
    if (costs)
    {
        size_t i;
        for (i = 0; i < size; ++i)
        {
            printf("%d\n", costs[i]);
        }
    }
    else
    {
        puts("Sorry, could not allocate memory.");
    }
    return 0;
}

int *initCosts(size_t size, int initial_value)
{
    int *costs = malloc(sizeof(*costs) * size);
    if (costs)
    {
        size_t i;
        for (i = 0; i < size; ++i)
        {
            costs[i] = initial_value;
        }
    }
    return costs;
}