Thread: double pointer problem

Hi, everyone.

I tried to handle my character pointer in the function but it's not easy..

At first, I allocate 3 lenths of character string and set as "EOF" out of function "strToHexa".
And then tried to make expand space to put 6 lengths of other strings with same character pointer by freeing and allocating memory again in the function. So I used double pointer as a parameter of function.

Code:

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

void strToHexa(char **string)
//convert characters to hexadecimal
{
    char toHexa[3];
    int i;
    int sLen=strlen(*string);//set the length of the original string into "sLen"
    printf("in strToHexa[before] : %p\n", *string);//result: "0090B928"
    if(sLen<=3){//if the length of original string is less than 3 (3 lengths is my limitation.
        strcpy(toHexa, *string);//copy original strings to arrays
        free(*string);//free original 
        *string=(char *)malloc(sizeof(char)*sLen*2);//allocate memory with 2 times of the original string because one character needs two characters to express hexadecimal
        for(i=0;i<sLen;i++){
            sprintf((*string)+(i*2), "%X", toHexa[i]); 
        }
    }
    printf("in strToHexa[after] : %p\n", *string); //result: "0090B928"
}


int main(void)
{
	char *test=(char *)malloc(sizeof(char)*3);
	strcpy(test, "EOF");
	printf("[before] %p\n", test); //result: "0090B928"
	printf("test: %s\n", test); // result: "EOF"

	strToHexa(&test);

	printf("[after] %p\n", test); //result: "FFFFFFFF"
	printf("test: %s\n", test); // segmatation fault! I wanned to see "454F46" which is hexadecimal form of "EOF"
    return 0;
}

Here is my another trying. It's simple and goes in right way.

Code:

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

void setStr(char **string)
{
	printf("in[before] %p\n", *string);
	free(*string);
	*string=(char*)malloc(sizeof(char)*4);
	strcpy(*string, "ABCD");
	printf("in[after] %p\n", *string);
}

int main(void)
{
	char *test=(char *)malloc(sizeof(char)*2);
	strcpy(test, "EO");
	printf("[before] %p\n", test);
	printf("test: %s\n", test);
	setStr(&test);
	printf("[after] %p\n", test);
	printf("test: %s\n", test);

    return 0;
}

The result is as follow.

[before] 009029B8
test: EO
in[before] 009029B8
in[after] 009029B8
[after] 009029B8
test: ABCD

Thank you so much, laserlight..!!
It's really helpful for me...

But still qurious why the seconde code works well.. even it has no '\0' character, either..

Originally Posted by Salem

> char *test=(char *)malloc(sizeof(char)*3);
> strcpy(test, "EOF");
It's already broken here, "EOF\0" is 4 characters, not 3.

The \0 trashes someone else's memory.

Sooner or later (anywhere between microseconds and years), your program will die with a segfault.

Your 'working' program has the same mistake, but that just makes you lucky rather than good.

OK, I can see clearly now with Salem's replayment...
I need to make my bad habit correct...

Thanks..!