Parameters and pointers with function inside function in C

Question

I'm having a bit of trouble trying to use pointers as parameters to a function inside another function. My goal is to preserve the value for the variable 'counter' in every function, in other words, when the lowest function declares a 'counter++', it's value must be incremented for every other 'counter' variable in program.

My code looks like this:

int main(int argc, const char * argv[]) {

    Hash tableHash;
    char command[6];
    int id = 0, counter = -1;
    int flags[4] = {0, 0, 0, 0};

    while(1) {
         identifyCommand(command, id, &tableHash, flags, &counter);
    }

    return 0;
    }

Inside my .h:

void identifyCommand(char* command, int id, Hash* tableHash, int* flag, int* counter){

    scanf("%s", command);

    /* ... */

    if(strcmp(command, "INSERT") == 0){
        scanf("%i", &id);
        commandInsert(id, tableHash, counter, flag);
    }

    /* ... */

    return;
}

void commandInsert(int id, Hash* tableHash, int* counter, int* flag){

    Registry x;
    x.key = id;

    if(flag[MALLOCFLAG]){
        tableHash->trees[*counter] = create_tree(x);
        counter++;
        flag[MALLOCFLAG] = 0;
    }
    else {
        insert_element(tableHash->trees[*counter], x);
    }
    return;
}

My main question is: when I run the code, it keeps sending the '-1' value of counter even after 'counter++' command is run inside commandInsert() function. Why is that happening and how do I solve it?

I think maybe the problem is at the commandInsert(id, tableHash, counter, flag) call, because I didn't use the reference sign (&), but 'counter' is already a pointer when inside identifyCommand() because of it's parameters, so what am I missing here?

P.P · Accepted Answer · 2015-12-08 23:11:52Z

6

Since you want to change the value pointed to by counter, you should change the value. But you are incrementing the pointer.

counter++;

should be

(*counter)++;

As @szczurcio noted, the command array you pass from can hold only upto 5 chars (1 for the NUL terminator). So the size of command array needs to be at least 7 to read "INSERT". To prevent buffer overrun, you can use the width in scanf() such as:

char command[7];
scanf("%6s", command);

Or you can use fgets().

char command[7];
fgets(command, sizeof command, stdin);

char *p = strchr(command, '\n');
if (p) *p = 0;

However, since you pass command to a function, you can't use sizeof command inside a function as that would return the sizoof(char*) (an array decays into a pointer to its first element when passed to a function). So you would have to pass the size information through another parameter:

From the caller:

   while(1) {
     identifyCommand(command, sizeof command, id, &tableHash, flags, &counter);
   }

And in the function:

void identifyCommand(char* command, size_t size, int id, Hash* tableHash,
 int* flag, int* counter){

   fgets(command, size, stdin);

   /* To remove the trailing newline, if any */
   char *p = strchr(command, '\n');
   if (p) *p = 0;

   ...

edited Dec 8, 2015 at 23:11

answered Dec 8, 2015 at 21:43

P.P

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5 Comments

StephenG - Help Ukraine Over a year ago

A useful convention when programming is to use a 'p' or 'ptr' at the end or start of a variable name to remind yourself it is a pointer and not a value. So in the OP's case use int *counterp as the parameter. This makes it easy to connect the dots mentally.

Mateus Hora Over a year ago

WOW, I totally didn't see that. Silly mistake that took me hours haha. Thanks a lot!!

P.P Over a year ago

@StephenG Indeed. At my work, there's a convention to use _p suffix for a pointer and _pp for a pointer to pointer. This way, one doesn't have to look up the declaration everytime and helps in accidents like this.

user4520 Over a year ago

It's worth mentioning that char command[6] cannot hold "INSERT", which appears to be one of the valid commands. And in general OP shouldn't use scanf without specifying the buffer width.

P.P Over a year ago

@szczurcio Thanks for pointing out. I have updated to cover that.

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