C struct pointers return

Question

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

#define MAX_BUFFER_SIZE 100   

typedef struct config_struct config_t ;

struct config_struct{
    char name[20];
        char ip[20];
        int port;
        char log[100];
};

int main(int argc, char *argv[]) {
        config_t config;
        read_config_file(argv[1], &config);
        return 0;
}

int read_int_from_config_line(char* config_line) {
        int val = atoi(config_line);
        return val;
}

void read_str_from_config_line(char* config_line, char* val) {
        strncpy(val,config_line, MAX_BUFFER_SIZE);
}

void read_config_file(char* config_filename, config_t *config) {
        FILE *fp;
        char *value, buffer[MAX_BUFFER_SIZE];
        fp = fopen("./settings.conf", "r");
        while (fgets(buffer, MAX_BUFFER_SIZE, fp) != NULL) {
            if(!strcmp(buffer,"NAME")) {
                    read_str_from_config_line(value, &(config->name));
        }   
            if(!strcmp(buffer,"PORT")) {
                &(config->port) = read_int_from_config_line(value);
                }

        }
}

I try to compile this code and it gives me a incompatible pointer type at read_str_from_config_line(value , &(config->name)); and lvalue required as left operand of assignment &(config->port) = read_int_from_config_line(value);

I am trying to return the stuct back to the main program but I'm having problem with the struct pointers.

Any solution for it to be solve? PS: The code is a mess as I trying to shorten it. Any help will be good as I am still a beginner to C programming.

Thanks

If you use array as left operand of = without [] operator it is automatically 'casted' into pointer to first element of the array. Try it without & operator, because it returns R value pointer to pointer value (I guess). — nikartix
– nikartix, Commented Mar 29, 2017 at 10:05
1) if(!strcmp(buffer,"NAME")) { --> if(!strcmp(buffer,"NAME\n")) { — BLUEPIXY
– BLUEPIXY, Commented Mar 29, 2017 at 10:07

unwind · Accepted Answer · 2017-03-29 10:07:06Z

2

The & operator means "take the address of the object to the right".

This:

&(config->port)

means "take the address of the port member of the structure pointed to by config.

You don't want to take the address, you just want to assign to the member. That's just:

config->port = ...

This is the same as (*config).port, but nobody writes it like that since the arrow operator is so convenient.

answered Mar 29, 2017 at 10:07

unwind

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davmac · Accepted Answer · 2017-03-29 10:13:50Z

2

                read_str_from_config_line(value, &(config->name));

confg->name is declared as a char array, which naturally decays to a pointer to the first element (of type "pointer-to-char") if you omit the &:

                read_str_from_config_line(value, config->name);

Taking its address explicitly instead results in a pointer of type pointer-to-array-of-20-chars, which is not the same type as pointer-to-char - which is why you are getting an error. This is a common source of confusion in C; in general, you should not take the address of an array (although it does have legitimate use cases).

In this line:

            &(config->port) = read_int_from_config_line(value);

You are taking the address of the port member of the structure pointed at by config, which is a non-lvalue pointer. It looks like you are actually trying to assign to the member, so the line should read:

            config->port = read_int_from_config_line(value);

edited Mar 29, 2017 at 10:13

answered Mar 29, 2017 at 10:07

davmac

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8 Comments

Jim Yong Over a year ago

alright thanks. what if i would like to printf config.name? How should i do it ?

Jim Yong Over a year ago

So in the whole code i'll use config->name instead of config.name for everything that i will do related will struct now?

davmac Over a year ago

@JimYong If you have a pointer to a struct, access its members using ->. So in your read_config_file function, use config->name, yes.

Jim Yong Over a year ago

ok thanks. problem now is atconfig->port = read_int_from_config_line(value); it's not reading to config->port.

davmac Over a year ago

@JimYong You are not intialising the value pointer that you are passing as an argument. Are you compiling with warnings enabled? You should have got a warning about that I think.

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