I am currently working on a project where I am manipulating command line data, but I can’t seem to get it work if I initialize the size of the array before.
char* len2[50]; // will store "1","2","3" from command line args
int size_arr = sizeof(len2) / sizeof(len2[0]);
printf("%d", size_arr);
this will input 50 when I am looking for it to input 3. How would I be able to find the size?
>Solution :
int size_arr = sizeof(len2) / sizeof(len2[0]);
sizeof(len2) asks for the total allocated size of len2 in bytes. This only works because C knows how many you allocated at compile time and turns it into a constant. It doesn’t tell you which ones you’ve filled in. C does not keep track of that; you have to.
Because len2 is an array of pointers, you can mark the end with a null pointer. The term for this is a sentinel value. First, be sure to initialize the array to null.
// This will initialize the entire array to NULL
char* len2[50] = {NULL};
Now you can find how many elements are in the array by looking for the first null, or by hitting the allocated size.
size_t len2_size = sizeof(len2) / sizeof(len2[0]);
int size = 0;
for( ; len2[size] != NULL && size < len2_size; size++ );
printf("size: %d\n", size);
This is, incidentally, basically how C strings work. The end of the string is marked with a 0.
Alternatively, you can store the allocated size and number of elements in a struct and keep track, but that’s another question.
Finally, if you’re just reading command line arguments, use argc and argv. argc is the size of argv. argv[0] is the name of the program, and the rest are the command line arguments.
int main( const int argc, const char **argv) {
printf("%d arguments in argv\n", argc-1);
}