C define array size automatically

Question

I'm trying to copy a const char* string to a char array[?]

is there a way in C to set my array size automatically? char word[255]; instead of using 255 my program will automatically use the correct size in my case 10. If there's a better way to do it I'm open to any suggestions thx a lot.

const char* test_str = "my string.";

   int lenght = strlen(test_str), x=0;
    char word[255] = {'\0'};
    //memset(word, '\0', sizeof(word));
    while (x < lenght) {
        word[x] = test_str[x];
        x++;
    }

    printf("%s", word);

edit: Removed memset(word, '\0', sizeof(word)); and replaced with word[255]= {'\0'};

In C99, you can use a VLA (variable length array) like this: char word[length + 1]; (+1 is necessary for the \0 terminator) — Felix G
– Felix G, Commented Jul 3, 2020 at 11:38
@FelixG True. Would not recommend it though. Better to look at malloc. — klutt
– klutt, Commented Jul 3, 2020 at 11:38
Read a good book about C programming, e.g. Modern C, then read this C reference and the documentation of your C compiler (perhaps GCC...) and debugger (perhaps GDB...) — Basile Starynkevitch
– Basile Starynkevitch, Commented Jul 3, 2020 at 11:46

RobertS supports Monica Cellio · Accepted Answer · 2020-07-03 12:37:28Z

3

You have two options here.

The first is using malloc. It's basically used like this:

char *word = malloc(sizeof *word * (strlen(test_str)+1));

and when you're done, you free the memory.

free(word);

The other alternative is using a VLA (variable length array):

char word[strlen(test_str) + 1];

I would recommend using malloc. It's a little bit messier, but in my opinion, VLA:s have considerable drawbacks, and if you're about to learn C, you will have to learn malloc sooner or later anyway, but you can do perfectly fine without VLA:s.

I have written an answer about why I consider VLA:s bad here: https://stackoverflow.com/a/58163652/6699433

edited Jul 3, 2020 at 12:37

RobertS supports Monica Cellio

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answered Jul 3, 2020 at 11:45

klutt

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6 Comments

mreff555 Over a year ago

I’m probably forgetting my order of operations but wouldn’t that allocate 40 bytes instead of 11? Why multiply by the pointer size?

mreff555 Over a year ago

*word is a char pointer. On modern systems that would be either 4 or 8 bytes depending on the architecture.

Nick S Over a year ago

sizeof *word == sizeof(char) and sizeof word == to the ptr size (4 or 8 bytes)

FatSumo Over a year ago

yes it worked!, I did without adding the +1 before but I figured it out and aded myself. thx a lot everybody. I did some programming courses before and my teacher told the I should always cast my malloc like this char* word =(char*) malloc(sizeof *word *strlen(text)+1);, can anyone tell me why ?. thx a lot.

RobertS supports Monica Cellio Over a year ago

@mreff555 In this case, the * operator is used to dereference the pointer to char word, not to declare word as pointer twice. sizeof *word is equal to sizeof (char).

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