集成終端和調試器產生不同的結果

Question

目標：打印得票最高的名字

所以我現在調試我的 PSET3 Plurality 大概有 8 個多小時了。 我認為一切都已修復，因為調試器中的結果是我想要的，但是，在我的集成終端中，它產生的結果不同。

調試器結果：

plurality/ $ debug50 plurality ./plurality me duck you

.

Number of voters: 7
Vote: me
Vote: me
Vote: me
Vote: d
Invalid vote.
Vote: duck
Vote: duck
Vote: you
me

綜合終端結果：

plurality/ $ ./plurality me duck you
Number of voters: 7
Vote: me
Vote: me
Vote: me
Vote: d
Invalid vote.
Vote: duck
Vote: duck
Vote: you
duck
plurality/ $ 

“我”應該贏了，但為什么“鴨子”贏了？

我的代碼（我想制作一個最小的可重現示例，但我不知道如何，這是我能想到的最小的，對不起。一直向下滾動。 ）：

#include <cs50.h>
#include <stdio.h>
#include <string.h>

// Max number of candidates
#define MAX 9

// Candidates have name and vote count
typedef struct
{
    string name;
    int votes;
}
candidate;

// Array of candidates
candidate candidates[MAX];

// Number of candidates
int candidate_count;

//Swapping Function
void swap (char ** xptr, char ** yptr);

// Function prototypes
bool vote(string name);
void print_winner(void);


int main(int argc, string argv[])
{
    // Check for invalid usage
    if (argc < 2)
    {
        printf("Usage: plurality [candidate ...]\n");
        return 1;
    }

    // Populate array of candidates
    candidate_count = argc - 1;
    if (candidate_count > MAX)
    {
        printf("Maximum number of candidates is %i\n", MAX);
        return 2;
    }
    for (int i = 0; i < candidate_count; i++)
    {
        candidates[i].name = argv[i + 1];
        candidates[i].votes = 0;
    }



    int voter_count = get_int("Number of voters: ");

    // Loop over all voters
    for (int i = 0; i < voter_count; i++)
    {
        string name = get_string("Vote: ");

        // Check for invalid vote
        if (!vote(name))
        {
            printf("Invalid vote.\n");
        }
    }

    // Display winner of election
    print_winner();
}

// Update vote totals given a new vote
bool vote(string name)
{
    // TODO
    for (int i = 0; i < candidate_count; i++)
    {
        if (strcmp(candidates[i].name, name) == 0)
        {
            candidates[i].votes += 1;
            return true;
        }
    }
    return false;
}

本節可能會幫助您...

// Print the winner (or winners) of the election
void print_winner(void)
{
    // TODO

    for (int z = 0; z < candidate_count - 1; z++)
    {
        int n = 0;
        while (n < candidate_count - 2)
        {
            int a = candidates[n+1].votes;
            int b = candidates[n+2].votes;
            char *x = candidates[n+1].name;        //me
            char *y = candidates[n+2].name;        //duck
            if (a > b)
            {
                a = a + b;                          //Swapping techniques credits to GeekforGreeks.org & Childre'sTechVideo
                b = a - b;
                a = a - b;
                swap(&x, &y);
            }
            candidates[n+1].votes = a;
            candidates[n+2].votes = b;
            candidates[n+1].name = x;
            candidates[n+2].name = y;
            n++;
        }
    }
    int c = 0;
    while (c != candidate_count - 1)
    {
        c++;
        if (candidates[candidate_count-1].votes == candidates[c].votes)
        {
            printf("%s\n", candidates[c].name);
        }
    }
}

//Swap function
void swap (char ** xptr, char ** yptr)
{
    char * tmp = NULL;
    tmp = *xptr;
    *xptr = *yptr;
    *yptr = tmp;
}

Answer 1

沒有必要嘗試排序...只需找到最高值並打印達到該數量的盡可能多的候選人。

int maxVotes = -1;
for (int z = 0; z < candidate_count; z++)
    if( candidates[z].votes > maxVotes )
        maxVotes = candidates[z].votes;

for (int z = 0; z < candidate_count; z++)
    if( candidates[z].votes == maxVotes )
        printf( "%s - %d votes\n", candidates[z].name, candidates[z].votes );