How to know how many strings does a string array contain?

RDP1998 0
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I've a string array:

char array[128][128];

To add strings in this array, I read a file, and append a line if "192.168.101.2" is contained in that line:

while ((read = getline(&line, &len, fp)) != -1) {
    if (strstr(line, "192.168.101.2") != NULL) {
        printf("%s\n", line);
        strcpy(array[i], line);
        i++;
    }
    
}

Now, I would like to know how many strings this array contains. I've tried: sizeof(array)/sizeof(array[0]), but it always returns 128. How can I achieve this? And, how can i pass an array of strings to a function? I've tried:

void array_length(char array[int][int]);

but:

main.c:15:31: error: expected expression before ‘int’
   15 | void array_length(char array [int][int]);
string arrays c
2 Answers

How to know how many strings does a string array contain?

C does not really have arrays, as they are in most other languages. Sure, you have array types, and can initialize arrays directly. But at runtime, array is just a pointer. There is no additional runtime information about it.

So, you have to keep track of both array maximum size and how many (or which) elements of array are used/initialized yourself.

So, here's code:

// char *array[128] might be better, but would require use of malloc for every string 
char array[128][128];

// array size in bytes / one row size in bytes = number of rows
const size_t array_capacity = sizeof (array) / sizeof(array[0]); 

// update len as you add strings of text 
size_t array_len = 0;
hyde0 
int length=0;
while(array[length][0]!='\0')
     ++length;
printf("%d",length); // in c/c++
Anukriti Gusain0
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