Difference between reading large chunks of data using char* or std::string

Question

I was trying to solve a coding problem and, to summarize, I was using this piece of code:

char result[50005][26], buffer[50005];
while (fin >> buffer) {
      for (int i = 0; i < strlen(buffer); ++i) {
          result[lin++][col] = buffer[i];
      }
      lin = 0;
      col++;
}

I needed to read some arbitrary words then store the letters. This worked and gave me 90 points, but it gave a time error. Then I switched the buffer from a char array to an std::string (and the strlen, of course), and it gave me 100 points.

Now I'm left dumbfounded because I can't really seem to grasp the difference between them two. Was reading the char array somehow also "reading" the remaining of the 50k imaginary bytes or something?

Why is strlen(buffer) repeatedly called? Once, before the loop is enough. A lessor compiler may not see that a repeated call is not needed. for (int i = 0; buffer[i]; ++i) { result[lin++][col] = buffer[i]; } would be enough. I think the switch to std::string afforded a O(1) time instead of strlen() needs O(n). — chux
– chux, Commented Apr 12 at 2:28
Side note: "I was trying to solve a coding problem" - be very careful. Most so-called "competitive coding" sites and sites saying they can increase your coding skills through solving their problems, are either huge scams or just teach you terrible coding practices and bad habits. Personally I'd stay away from those sites. — Jesper Juhl
– Jesper Juhl, Commented Apr 12 at 16:11

Ryan Zhang · Accepted Answer · 2025-04-12 02:52:11Z

5

See the "Possible implementation" of strlen in the C++ docs. Note that not only is O(1) performance not guaranteed, it would be impossible to achieve. strlen only receives an array of char, and hence has no clue how long your string is; it must iterate until the end of the string is reached. On the other hand, std::string has a size_t member that keeps track of the string length, as listed in the docs. Hence reading its length with .length() is O(1).

Hence, in your implementation, strlen is called in each iteration of the loop, so the total time complexity is O(N) per loop of N iterations = O(N^2). Whereas, using .length() would be O(1) per loop of N iterations = O(N).

A simple fix would be to read the length of buffer once prior to the loop and store it, for example with size_t len = strlen(buffer) . Or, you may iterate buffer until the end of the string by checking if buffer[i] is '\0'.

answered Apr 12 at 2:52

Ryan Zhang

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1 Comment

naphy Apr 12 at 16:07

Once again, I keep falling for this stupid mistake. In my head, I can't fathom that getting the length of a string of size 4 would take that much of a strain. But here I am, proven wrong once more.

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