输入C样式的字符串并获取长度

Question

The string input format is like this

str1 str2

I DONT know the no. of characters to be inputted beforehand so need to store 2 strings and get their length. Using the C-style strings ,tried to made use of the scanf library function but was actually unsuccessful in getting the length.This is what I have:

// M W are arrays of char with size 25000
   while (T--)
   {
       memset(M,'0',25000);memset(W,'0',25000);
       scanf("%s",M);
       scanf("%s",W);
       i = 0;m = 0;w = 0;
       while (M[i] != '0')
              {
                  ++m; ++i;  // incrementing till array reaches '0'
              }
        i = 0;
       while (W[i] != '0')
              {
                  ++w; ++i;
              }
       cout << m << w;


   }

Not efficient mainly because of the memset calls.

Note: I'd be better off using std::string but then because of 25000 length input and memory constraints of cin I switched to this.If there is an efficient way to get a string then it'd be good

Answer 1

Aside from the answers already given, I think your code is slightly wrong:

   memset(M,'0',25000);memset(W,'0',25000);

Do you really mean to fill the string with the character zero (value 48 or 0x30 [assuming ASCII before some pedant downvotes my answer and points out that there are other encodings]), or with a NUL (character of the value zero). The latter is 0 , not '0'

   scanf("%s",M);
   scanf("%s",W);
   i = 0;m = 0;w = 0;
   while (M[i] != '0')
          {
              ++m; ++i;  // incrementing till array reaches '0'
          }

If you are looking for the end of the string, you should be using 0 , not '0' (as per above).

Of course, scanf will put a 0 a the end of the string for you, so there's no need to fill the whole string with 0 [or '0' ].

And strlen is an existing function that will give the length of a C style string, and will most likely have a more clever algorithm than just checking each character and increment two variables, making it faster [for long strings at least].

Answer 2

You do not need memset when using scanf , scanf adds the terminating '\\0' to string.

Also, strlen is more simple way to determine string's length:

scanf("%s %s", M, W); // provided that M and W contain enough space to store the string
m = strlen(M); // don't forget #include <string.h>
w = strlen(W);

Answer 3

C-style strlen without memset may looks like this:

#include <iostream>
using namespace std;

unsigned strlen(const char *str) {
    const char *p = str;
    unsigned len = 0;
    while (*p != '\0') {
        len++;
        *p++;
    }
    return len;
}

int main() {
    cout << strlen("C-style string");
    return 0;
}

It's return 14.