another strlen

Hi people! I'm learning C now and hope someone can help me with the task. I thought I would be able to write strlen function but it doesn't work. Here is the code:

Code:

#include "mystrlen.h"

int mystrlen(char *s){
	
	int i,k;
	for(i = 0; s[i] != '\0'; i++){
		k++;
	}
	
	return k;
}

Code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "mystrlen.h"

int main () {
   
    int i;
	char *p1 = "Use malloc()";
	char *p2;
	
	if (NULL == (p2 = malloc(mystrlen(p1) + 1))){
		printf("Not enough memory!\n");
		return 1;
	}
	
	for (i = 0; i <= mystrlen(p1); i++)
		p2[i] = p1[i];
	
	printf("%s\n",p2);
	free(p2);

    return 0;
}

You need to initialize k, although you could just use i instead and get rid of k completely.

Code:

#include "mystrlen.h"

int mystrlen(char *s){

	int i;
	
	for(i = 0; s[i] != '\0'; i++){
	}
	
	return (i);

}

Running…
0 0 0 0 0 0 0 0 0 0 //normally it should print Use malloc() here
Debugger stopped.
Program exited with status value:0.

That is very weird...you sure you didn't forget to build the project or anything like that? Because with your original source and with your modified strlen everything was working fine.

can it be because of mac ? compiler maybe

Mac should not matter, and I doubt you have found a bug in the compiler. Either you are running the wrong executable, you are compiling the wrong code (or not compiling at all) or the source you posted here is not in line with the source you currently have.

You do need to manually add the null terminator to the end of the string in p2 after your loop, but that's not why you are getting 0s printed.

Actually he doesnt , notice how he loops to i <= strlen, which means he will get the trailing \0 when he does the string copy.

Yep. Thanks for the correction.

big thanx all for help! I woke up today with the idea to reorganize things in the code and it's strange but now it's printing Use malloc() like it should )

Code:

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

int mystrlen(char *s){
	
	int i;
	for(i = 0; s[i] != '\0'; i++){
	}
	
	return i;
}


int main () {
   
    int i;
	char *p1 = "Use malloc()";
	char *p2;
	
	if (NULL == (p2 = malloc(mystrlen(p1) + 1))){
		printf("Not enough memory!\n");
		return 1;
	}
	
	for (i = 0; i <= mystrlen(p1); i++)
		p2[i] = p1[i];
	
	printf("%s\n",p2);
	free(p2);
	
    return 0;
}

Code:

Loading program into debugger…
Program loaded.
run
[Switching to process 1516]
Running…
Use malloc()

Debugger stopped.
Program exited with status value:0.

I added mystrcpy function which is replacing strcpy and it returns null, somebody knows why?

Code:

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

int mystrlen(char *s){
	
	int i;
	for(i = 0; s[i] != '\0'; i++){
	}
	
	return i;
}

void mystrcpy(char *s1, char *s2){
	
	int i;
	
	if (NULL == (s2 = malloc(mystrlen(s1) +1))){
	   printf("Not enough memory!\n");
	}
	
	for (i=0; s2[i] != '\0' ; i++)
		s2[i] = s1[i];
}

int main () {
   
	char *p1 = "Use malloc()";
	char *p2;
	
	mystrcpy(p1, p2);
	
	printf("%s\n",p2);
	free(p2);
	
	return 0;
}

This GDB was configured as "x86_64-apple-darwin".tty /dev/ttys000
Loading program into debugger…
Program loaded.
run
[Switching to process 2082]
Running…
(null)

Debugger stopped.
Program exited with status value:0.

Notice that mystrcpy changes the parameter named s2. It does not change the local variable in main named p2. One option is to use a pointer to a pointer.

no idea how to do this

Question 4.8

Aside from the s2 pointer getting lost as soon as the function returns that others have mentioned, there's also:

Code:

	for (i=0; s2[i] != '\0' ; i++)
		s2[i] = s1[i];

I think you mean:

Code:

	for (i=0; s1[i] != '\0' ; i++)
		s2[i] = s1[i];

What you're really trying to recreate is strdup(). You can see, the copied-to memory is simply returned by strdup(), because it's easier than having you pass in a pointer to a pointer to the allocated memory.