Copying float values into char array

Question

I'm writing a TCP socket in C to send location data for a project I'm working on.

So far, everything works, but I'm struggling with this seemingly simply problem. I'm trying to build a JSON String that will be sent over the socket. I have a character array (representative of the String) json defined as:

char json[1024];

With a method prototype:

const char* build_json(void);

And method body:

const char* build_json(void) {
    strcpy(json, "{");
    strcat(json, "\"latitude\":");
    sprintf(json, "%0.5f", latitude);
    strcat(json, "}");
    return json;
}

I know that latitude is defined correctly and should be a float of approximately 5 decimal places.

But when I call build_json(); , 38.925034} is the only thing that is returned. Why is this the case? It appears that the call to sprintf is overwriting what's already been written in json .

Thanks for your help!

Answer 1

sprintf will not append to your string; rather, it will overwrite whatever is there. You could do this:

sprintf(json + strlen(json), "%0.5f", 213.33f);

But, to be honest, this is a much better solution:

sprintf(json, "{\"latitude\":%0.5f}", location);

And this solution is still better:

snprintf(json, sizeof(json), "{\"latitude\":%0.5f}", location);
json[sizeof(json) - 1] = '\0';

as long as json is an array visible to the function that calls snprintf , ie allocated in that function on the stack, or globally. If it's a char* that you pass to the function, this will fail miserably, so beware.

Answer 2

You had better to do this only with sprintf to avoid multiple operations.

const char* build_json(void) {
    sprintf(json, "{\"latitude\":%0.5f}", latitude);
    return json;
}

Moreover if you are writing network code, you had better to allocate your string in your function and not relying on global. Often, network code are done in a multi-thread way.

Answer 3

Quick & dirty fix:

char* cptr = json;
...
strcpy(cptr, "{");
cptr += sizeof("{") - 1;

strcat(cptr, "\"latitude\":");
cptr += sizeof("\"latitude\":") - 1;

sprintf(cptr, "%0.5f", latitude);

The proper solution would be put the string literals and their sizes in constant variables instead of the above.

char* cptr = json;
...
strcpy(cptr, STR_START_BRACE);
cptr += STR_START_BRACE_LEN;

strcat(cptr, STR_LATITUDE);
cptr += STR_LATITUDE_LEN;

sprintf(cptr, "%0.5f", latitude);

Answer 4

You may write the whole string at once in buffer:

const void build_json(char * json, size_t *len) 
{
    char buff[50];
    sprintf(buff, "{\"latitude\":%0.5f}", latitude);
    *len = strlen(buff);
    strncpy(json, buff, *len);
}

Just provide enough space for buffer, you need to allocate your json outside of the func and free it when out of scope