无法将传递数组理解为命令行参数

Question

/*

I am trying to program a Tideman alternative method algorithm.
I am taking i/p through command-line-argument.
When I try to print the populated candidate array.
It is printing single characters when format specifier is char type. The result of other specifier is attached.
*/

/*

Same logic when executed on CS50 IDE, it prints the strings as expected.
I mean if the CLA is Alice Bob Charlie, then at 0th index Alice and so on...
This step is crucial as later the names from the voters would be compared.
I don't how to proceed from here. I came to know only about this when I reached the comparison stage. */

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

#define MAX 9

int preferences[MAX][MAX]; 


bool locked[MAX][MAX];

typedef struct
{
    int winner;
    int loser;
}
pair;

char candidates [MAX] = {0};
pair pairs[MAX*(MAX-1)/2];


int pair_count = MAX;
int candidate_count = MAX;

//Prototypes
bool vote (int rank, char name, int ranks[]);
void record_preferences(int ranks[]);
void add_pairs();
void sort_pairs();
void lock_pairs();
void print_winner ();

int main(int argc, char *argv[])
{

//Check for invalid usage.  
    if (argc < 4)
    {
        printf("Usage: tideman [candidate ...]\n");
        return 1;
    }

    candidate_count = argc - 1;
    if (candidate_count > MAX)
    {
        printf("Maximum number of candidates is %i\n", MAX);
        return 2;
    }
//popuplate the candidate array.  
    for (int i = 0; i < candidate_count; i++)
    {
        candidates[i] = argv[i+1];
        printf("Candidate array: %c\n", *argv[i+1]);
    }
}

Answer 1

Your problem is understanding how arguments work. Consider this line in the terminal: ./main tideman abc How does argv looks like? It is an array.. with 3 elements in it:

argv[0] => "main"
argv[1] => "tideman"
argv[2] => "abc"

Lets say instead of abc you have ./main tideman ab c . How does argv looks like? It is an array.. with 5 elements in it:

argv[0] => "main"
argv[1] => "tideman"
argv[2] => "a"
argv[3] => "b"
argv[4] => "c"

candidates is a char array, so it can hold characters. Now lets consider first example ( ./main tideman abc ). What is the for-loop in your code does? candidates[0] = argv[1] . What is wrong here? candidates[0] is the first character in the candidates array.. Cool. argv[1] is "tideman" .. a string. I hope you can see the assignment problem here.

Now the second problem, in the second part of the loop. With the first iteration, argv[0+1] is "tideman" Now you want to print the string tideman to the screen. The command print("%s\n") does print a string, and it take as argument pointer to that string. What is the pointer? its argv[1] .. which is a pointer to the first element in the argv[1] array. What is *argv[1] ? Thats right, its the value of the first character in the argv[1] array.. So in our example argv[1] is "tideman", so *argv[1] is 116 which is the ascii value for t .

So in order to achieve your goal of printing "tideman" , which is argv[1] you just need printf("%s\n", argv[1]); with argument argv[1] you supply pointer to the first character of the string, which is what %s ask for.

So to recap you have 2 problems in your for-loop, and you could see both of those problem as warnings in the terminal.

Answer 2

In C language, a char is a single character. A string is by convention a null terminated character array. For that reason, it is common to use char * to represent strings. And it is up to the programmer to make sure that the life time of the pointed array does not end before the value is used.

Warnings are to be observed. Unless you know what a warning means and why you want to ignore it, you really should treat them as errors.

Here the information was in the first warning you have shown:

    candidates[i] = argv[i + 1];

... warning: assignment to char from char *...

It means that argv[i+1] does represent a string, but you try to assign it into a single char which invokes Undefined Behaviour.

Here is a quick fix for your current code:

...
char *candidates [MAX] = {NULL};  // candidates is an array of strings
...
    //popuplate the candidate array.  
    for (int i = 0; i < candidate_count; i++)
    {
        candidates[i] = argv[i + 1];
        printf("Candidate array: %s\n", argv[i + 1]);    // use %s in C to display a string
    }
...