how to assign a string to a fixed sized char array

Question

I am reading a string from a json object. The string can be compressed or uncompressed. If it's compressed, I have to decompress it. So depending on the compressed condition I want to assign a value to with json_string_value. I know the size of the string, hence I want the string to have a static size.

I have the following:

char my_string[MY_SIZE];

if( [some condition]){ 
    //how to assign a value to my_string in this case?
} else {
    ...
    int ret = decompress(compressed_str, compressed_str_len, my_string, MY_SIZE);
    ...
}

json_string_value() returns a string with a null terminator.

I managed to get it to work by using a different string literal and copying the value over

const char *tmp = json_string_value(image);
strcpy(my_string, tmp);

but I was wondering if there is an easier (better) way to do this?

Where do you have a string literal? String literals are in double quotes, e.g. "foo" — Barmar
– Barmar, Commented Feb 16, 2022 at 23:32
@Barmar: right, sorry, mixed up names. And sorry again, used the wrong variable names, edited my answer — Tom
– Tom, Commented Feb 17, 2022 at 0:29
Slightly off-topic: beware buffer overrun when you use strcpy. Consider strncpy instead. — Robᵩ
– Robᵩ, Commented Feb 17, 2022 at 0:33

Barmar · Accepted Answer · 2022-02-16 23:33:28Z

2

You don't need all the extra variables, you can just call the function in strcpy() arguments.

strcpy(my_string, json_string_value(image));

answered Feb 16, 2022 at 23:33

Barmar

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3 Comments

pm100 Over a year ago

" I know the size of the string" I would still strncpy just in case, cant be too careful in C

Barmar Over a year ago

If you use strncpy() and the length is too short you end up with an unterminated string. It prevents buffer overflow, but causes other problems. It's not a magic bullet.

pm100 Over a year ago

TIL that strncpy_s is part of C11. So thats what should be used

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