double pointer problem

This is a discussion on double pointer problem within the C Programming forums, part of the General Programming Boards category; Hi, everyone. I tried to handle my character pointer in the function but it's not easy.. At first, I allocate ...

  1. #1
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    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

  2. #2
    C++ Witch laserlight's Avatar
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    I note that "EOF" consists of 4 characters: 'E', 'O', 'F' and '\0'. As such, it cannot fit into an array of 3 characters while remaining a string which can be used with string functions like strlen().
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  3. #3
    and the hat of wrongness Salem's Avatar
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    > 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.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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  4. #4
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    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..

  5. #5
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    Quote Originally Posted by Salem View Post
    > 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..!

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