C parsing a string into an array of strings

Question

While defining an array of strings, I usually declare it similar to the following:

    char *arr[5] =
{
    "example0",
    "example1",
    "example2",
    "example3",
    "example4"
};

Where I'm having a problem is I don't know how to pass a variable into one of the elements of arr.

For instance,

char str[6] = "1.0.0.1";

char *arr[6] =
{
    "example0",
    "example1",
    "example2",
    "example3 %s", str,
    "example4"
};

Of course, this doesn't work, it's just a basic illustration of what I'm having trouble with.

I also know I can later use strncat() or even snprintf() but, to avoid the pain of handling memory with those, I just want to know if parsing a variable into one of the strings of the array is possible at declaration.

The posted code would "work" if str were sufficiently sized for the initialization value you're providing. By "work" I mean compile. Obviously It isn't going to magically plant str into the example3 %s string at the apparent format specifier location. — WhozCraig
– WhozCraig, Commented Apr 16, 2022 at 20:28
Ah, that magic is what I need. That code compiles anyways, by doesnt work I meant that magic doesnt happen. Also, 6 is enough for handling str as another element. — Ömer Gök
– Ömer Gök, Commented Apr 16, 2022 at 20:31
I believe in C++, you can do something like: "bla bla text " + str + "bla bla text". Nothing like that in C? — Ömer Gök
– Ömer Gök, Commented Apr 16, 2022 at 20:32
This isn't C++, and no, you don't get that functionality in C. And read what I said again. I didn't say arr wasn't sufficiently sized. I said str is not. The initialization literal "1.0.0.1" you're providing requires 8 chars of storage; you're specifically declaring str as char[6]. 8 doesn't fit into 6, no matter how hard you push. — WhozCraig
– WhozCraig, Commented Apr 16, 2022 at 20:50
The undersized array I have pointed out now twice is str, not arr If you removed literally every line from you program except char str[6] = "1.0.0.1"; the problem I'm describing would remain. str is undersized. You cannot shove 8 chars into a char buffer declared to be 6. If there is a way I can say that differently than I already have, it eludes me. And the sizeof of str is not added to arr[3]. The array arr holds six char pointers, not char arrays. The fifth element in that array, arr[4], is the base address of str. — WhozCraig
– WhozCraig, Commented Apr 16, 2022 at 21:10

chux · Accepted Answer · 2022-04-17 11:12:59Z

3

... if parsing a variable into one of the strings of the array is possible at declaration.

At compile time, could concatenate as below:

#define STR "1.0.0.1"
char str[] = STR;

char *arr[6] = { 
    "example0",
    "example1",
    "example2",
    "example3" " " STR, // Forms "example3 1.0.0.1"
    "example4"
};

Perhaps OP is interested in something formed during run-time. It uses a variable length array (VLA).

void foobar(const char *str) {
  int n = snprintf(NULL, 0, "example3 %s", str);
  char a[n]; // VLA.
  snprintf(a, sizeof a, "example3 %s", str);

  char *arr[6] = {
      "example0",
      "example1",
      "example2",
      a,
      "example4"
  };

  printf("<%s>\n", arr[3]);
}

int main(void) {
  foobar("1.0.0.1");
}

Output

<example3 1.0.0.>

Alternatively the space for the string could have been done via an allocation.

char *a = malloc(n + 1u);
sprintf(a, "example3 %s", str);
....
free(a);

edited Apr 17, 2022 at 11:12

answered Apr 16, 2022 at 20:41

chux

158k17 gold badges160 silver badges311 bronze badges

Sign up to request clarification or add additional context in comments.

11 Comments

Ömer Gök Over a year ago

I really don't understand why you used char str[] in your example. That is a string constant not a variable.

Robert Harvey Over a year ago

@ÖmerGök A variable would not work at compile time. You would need something like sprintf("example3 %s", str); which only works at runtime.

WhozCraig Over a year ago

^^^ including the target buffer, obviously.

chux Over a year ago

@ÖmerGök char str[] = STR; is illustrative code to show how str and arr can both derive their initializations from STR. str is not needed to initialize arr.

Ömer Gök Over a year ago

snprintf(arr[3], MAX, "example3 %s", str) wouldn't work anyways. At this point, I think I should just allocate sufficient memory before doing anything.

|

Collectives™ on Stack Overflow

C parsing a string into an array of strings

1 Answer 1

11 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

11 Comments

Your Answer

Sign up or log in

Post as a guest

Related