C编程找到2维数组的长度

Question

My program grabs command line arguements with argc and argv[]. My question is how can I find the length of argv[1][i].

My code that grabs length of argv[]

int my_strlen(char input[]){ 
    int len = 0;
    while(input[len] != '\0'){
       ++len;
    }
    return len;
}

but when I try to find argv[1][len] I get a subscripted value is neither array nor pointer: my attempt

int my_strlen(char input[]){
   int len = 0;
   while((input[1][len] - '0') != '\0'){
      ++len;
   }
   return len;
}

FULL CODE:

#include <stdio.h>
#include <math.h>

int my_strlen(char input[]);

int main(int argc, char *argv[]){
int length = 0;
length = my_strlen(argv[1]);
long numberArr[length];
int i, j;

for(i = 0; i < length; i++){
  numberArr[i] = argv[1][i] - '0';
}

 return 0;
}

int my_strlen(char input[]){
 int len = 0;
 while((input[1][len] - '0') != '\0'){
  ++len;
 }
 return len;
}

Thanks for any help in advance!

Answer 1

I think you're confused about the argv content. The OS will pass a number of ASCIIZ strings, such that invoking my_program with arguments ala...

my_program first second third

...is similar to having the following declaration in your program...

int argc = 4;
const char* argv[4] = { "my_program", "first", "second", "third" };

Hence, when you index into argv[1][i] you're getting the i-th character in the string "first". That's only valid for values of i between 0 (which yields 'f'), and 5 (which indexes to the terminating NUL character '\\0').

So, there no two-dimensional N*M array, but there is an array of pointers-to-(array-of-char). You can invoke the normal strlen() function as in strlen(argv[1]) to find out the number of characters in each argument. Only argc tells you the total number of elements in argv .

Does that help?

Answer 2

In main , you're passing argv[1] to my_strlen . That means my_strlen just receives a normal, single-dimension string. It doesn't need to do input[1][len] , just input[len] .