C error in output manipulating arrays through a function

Question

Well, I have written this simple program that, from a char array and a keyword, stores in another array the part of the string that begins with that keyword.

It works if i don't use any function to manipulate the array, but what i don't understand is why it doesn't work when it's manipulated through a function, there is no output from printf in this case.

Code that don't work:

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

void createRequest(char *buffer) {
    strcat(buffer, "GET /vfolder.ghp HTTP/1.1\r\n");
    strcat(buffer, "User-Agent: Mozilla/4.0\r\n");
    strcat(buffer, "Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8\r\n");
    strcat(buffer, "Conection: Keep-Alive\r\n\r\n");
    strcat(buffer, "name=2&password=3");
}

void getParameters(char *buffer, char *parameters) {
    for (int i = 0; i < sizeof(buffer); i++) {
        if (strncmp(buffer + i, "name=", 5) == 0) { strcpy(parameters, buffer + i); }
    }
}

int main(int argc, char *argv[]) {
    char buffer[250];
    char parameters[250];

    memset(buffer, 0x00, sizeof(buffer));
    memset(parameters, 0x00, sizeof(parameters));

    createRequest(buffer);

    fprintf(stdout, "Your buffer is: \n%s\r\n", buffer);

    getParameters(buffer, parameters);

    printf("The parameters are: \n%s\r\n", parameters);
}

Code that work:

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

void createRequest(char *buffer) {
    strcat(buffer, "GET /vfolder.ghp HTTP/1.1\r\n");
    strcat(buffer, "User-Agent: Mozilla/4.0\r\n");
    strcat(buffer, "Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8\r\n");
    strcat(buffer, "Conection: Keep-Alive\r\n\r\n");
    strcat(buffer, "name=2&password=3");
}

int main(int argc, char *argv[]) {
    char buffer[250];
    char parameters[250];

    memset(buffer, 0x00, sizeof(buffer));
    memset(parameters, 0x00, sizeof(parameters));

    createRequest(buffer);

    fprintf(stdout, "Your buffer is: \n%s\r\n", buffer);

    for (int i = 0; i < sizeof(buffer); i++) {
        if (strncmp(buffer + i, "name=", 5) == 0) { strcpy(parameters, buffer + i); }
    }

    printf("The parameters are: \n%s\r\n", parameters);
}

Does this answer your question? Why isn't the size of an array parameter the same as within main? — Adrian Mole
– Adrian Mole, Commented Apr 25, 2020 at 18:32

MrBens · Accepted Answer · 2020-04-25 16:49:37Z

1

You shouldn't use sizeof in getParameters function.

The value of this operation is sizeof(char*), and this is not what you meant.

Change it and use strlen instead:

void getParameters(char *buffer, char *parameters) {
    for (int i = 0; i < strlen(buffer); i++) {
        if (strncmp(buffer + i, "name=", 5) == 0) { strcpy(parameters, buffer + i); }
    }
}

Basically, because it is a char* (which is null-terminated), you can use strlen in order to find its length. For an array of integers, for example, you will have to pass the numbers of elements as an extra parameter.

edited Apr 25, 2020 at 16:49

answered Apr 25, 2020 at 16:17

MrBens

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5 Comments

john Over a year ago

To be strictly accurate, they shouldn't have used sizeof in the 'working' code either. It's wrong in both cases, just more drastically wrong in the second case.

MrBens Over a year ago

@john Absolutely.

prosold Over a year ago

It's true i didn't realize that! Thank you very much it is working now, but, in the second case why sizeof is working in main function and is giving me the length of the array instead of giving me the size of char*?

MrBens Over a year ago

Well, because it is not char* :) It's an array of chars sitting on the stack.

MrBens Over a year ago

The subject of pointers and arrays in C is not easy to grasp, but it's a crucial step towards understanding C. Once you get hold of it, you begin to unleash the full power of the language. Start here: stackoverflow.com/questions/1335786/…

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