Re: C++ programming challenge

Jonathan Lee <>
Sat, 13 Jun 2009 10:26:00 -0700 (PDT)
Here's a submission in C++. Nothing that interesting, but to
illustrate a
point: the speed of this program is dominated by the printing of the
floating point value representing percent. i.e., this line:

    cout << ("abcdefghijklmnopqrstuvwxyz")[i] << ' ' << p <<"%\n";

if you remove "<< p" it will only take 0.158s on my computer. With
that in, it takes
0.514s. Perhaps it's just g++'s implementation of cout's float to text
In any event, I think we're at the mercy of the compiler's libs.

For what it's worth I also coded up a version that read longs instead
of chars, and
accumulated 16 bits at a time in the "count" array (fixing the
calculation at the
end). This didn't make much of a difference, though, in light of the
cout penalty
noted above.


#include <cstdio>
#include <iostream>
using ::std::cout;
using ::std::endl;

#define BUFFSIZE 4096

int main(int argc, char* argv[]) {
    if( argc <= 1 ) return (printf("Please provide a file name\n"), 1);
    CStopWatch s;

    if (FILE* f = fopen(argv[1], "r")) {
        unsigned char buff[BUFFSIZE];
        unsigned long count[256] = {0};
        unsigned long total_alpha[26] = {0};

        size_t bytesread;

        while ((bytesread = fread(buff, 1, BUFFSIZE, f)) > 0) {
            for (size_t i = 0; i < bytesread; ++i) {

        unsigned long total = 0;
        for (int i = 0; i < 26; ++i) {
            unsigned long x;
            x = count[("abcdefghijklmnopqrstuvwxyz")[i]]
              + count[("ABCDEFGHIJKLMNOPQRSTUVWXYZ")[i]];
            total_alpha[i] = x;
            total += x;
        float p2 = 100.0f / total;
        for (int i = 0; i < 26; ++i) {
            float p = p2 * total_alpha[i];
            cout << ("abcdefghijklmnopqrstuvwxyz")[i] << ' ' << p <<"%\n";

        cout << endl;
    cout << "Time elapsed: " << s.getElapsedTime() << endl;

    return 0;

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