Parsing C string and multiple pointer c string

Question

I'm trying to make a parser. I have a version working with just pointers but I want to know how to convert it from pointers (char *p) to c strings (char p[]) to get more familiar with how they work.

I have a function...

int makearg(char s[], char**args[])
{
    char *arg1
    char arg2[50];

    //I'm aware these don't assign the args pointer value to arg2.
    arg2[0] = args[0];

    //And I'm aware this does.
    arg1 = (*args)[0]);
}

My question is what is the equivalent. Or maybe a brief explanation on what's going on. Other solutions just give working code without really saying how it works.

I don't think this is related by my mem allocation is using the pointer and []. And that works fine. Little unsure how that works also if anyone has time.

(*args)[i] = (char*) malloc(sizeof(char) * tokenLen);

Welcome to StackOverflow! Which compiler error are you asking about? What words, specifically, from those errors have you confused? When you post examples here you need to be certain they produce only the problem you're asking about, and no other problems. You can do this by producing an [MCVE](http://stackoverflow.com/help/mcve). — autistic, Feb 07 '17 at 02:04
Invalid conversion from char* to char** or whatever combination I was trying. I knew the types were different but I didn't understand how. Every time I compiled it would give me an error and then I would adjust to get the error to go away but the pointer would be wrong. So I think I get it now. Thanks so much. — cba1067950, Feb 07 '17 at 10:16

score 1 · Accepted Answer · edited May 23 '17 at 12:33

The first statement:

arg2[0] = args[0];

arg2[0] is of type char and args[0] is of type char**. This assignment wont work because they are of different types.

However, something like this would work:

arg2[0] = (**args)[0]; /* or even (*args)[0][0] */

because you are giving arg2[0] a char, which is correct because they are matching types. **args is a pointer to a pointer to the first element of args, which is char*, and to reach a char, you need to index [0]. Their are multiple ways you could do this. Assigning arg2[0] to args[0][0][0] or even ***args is valid, even though these assignments would look obscure.

The second statement:

arg1 = (*args)[0];

will work because you are giving arg1, a char* pointer.*args is a pointer to first element of of args, which is char**, and index[0] locates to a char* pointer. This stores the address of (*args)[0]; within arg1. This could even be written in a much less popular way &(*(*args)[0]), although this is harder to read and not necessary.

Additionally, Your memory allocation is fine. (*args)[i] is a pointer to the first element of args, which is char**, and [i] would reference a char* pointer. You would need to allocate for char** before hand, as this is also an uninitialized pointer to pointer.

You can also shorten:

(*args)[i] = (char*) malloc(sizeof(char) * tokenLen);

to:

(*args)[i] = malloc(tokenLen);

As you don't need to cast return of malloc(), and sizeof(char) is always 1. '

You are a saint. Thank you heavenly spirit. <3 – cba1067950 Feb 07 '17 at 09:47 — cba1067950, Feb 07 '17 at 09:47
@cba1067950 No problem :) I'm glad I could help. – RoadRunner Feb 07 '17 at 10:32 — RoadRunner, Feb 07 '17 at 10:32

score 0 · Answer 2 · answered Feb 07 '17 at 00:35

char[] and char * is effectively the same thing because the compiler converts arrays to pointers. So there is no need to convert between them.

Think of the array syntax as a shortened version of a pointer dereference with an offset: arg[5] == *(arg + 5)

arg2[0] = args[0]; is not going to compile because args[0] is char** and arg2[0] is of type char.

arg1 = (*args)[0]); works because you they both have the same type - char *.

Parsing C string and multiple pointer c string

2 Answers2