我如何在（int argc，char ** argv）中打印f argv

Question

我想知道如何才能打印argv的特定部分。 例如

./program hello world
The first 3 letters of the first argument is:
1 = h
2 = e
3 = o
The full argument was: hello

The first 3 letters of the second argument is:
1 = w
2 = o
3 = r
The full argument was: world

这是我的代码，我只是不知道该放在那个地方

int main(int argc, char **argv) {
    printf("The first 3 letters of the first argument is:\n");
    int i = 0;
    while (i < 3) {
        printf("%d = %c\n", .... I don't know what to put here);
        i++;
    }
    printf("The full word was: %s\n\n", I don't know what to put here);

    printf("The first 3 letters of the second argument is:\n");
    int j = 0;
    while (j < 3) {
        printf("%d = %c\n", j, .... I don't know what to put here);
        j++;
    }
    printf("The full word was: %s\n", I don't know what to put here);
}

Answer 1

第一个参数在argv[1] ，第二个参数在argv[2] ，依此类推。 要从中打印单个字符，请添加另一级下标。

您还应该在打印参数之前检查是否已提供参数，并且参数是否有足够的字符可在循环中打印。

int main(int argc, char **argv) {
    if (argc >= 2) {
        printf("The first 3 letters of the first argument is:\n");
        int i = 0;
        while (i < 3 && argv[1][i]) {
            printf("%d = %c\n", i, argv[1][i]);
            i++;
        }
        printf("The full word was: %s\n\n", argv[1]);
    }

    if (argc >= 3) {
        printf("The first 3 letters of the second argument is:\n");
        int j = 0;
        while (j < 3 && argv[2][j]) {
            printf("%d = %c\n", j, argv[2][j]);
            j++;
        }
        printf("The full word was: %s\n", argv[2]);
    }
}

Answer 2

argv中的第一个元素是程序的名称，其余元素指向参数。 每个参数的前三个字符可通过遍历参数argv [1]至argv [n]来找到，同时为每个参数打印所需的字符。

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

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

    printf("Program:    %s\n", argv[0]);

    if (argc > 2)
    {
        int i = 1;
        for ( ; i < argc ; i++)
        {
            if (!argv[i][2])
            {
                fprintf(stderr,"Length of argument \"%s\" is less than 3\n", argv[i]);
                return(EXIT_FAILURE);
            }
            else
                printf("The first 3 letters of argument %d is:\n", i);
                int j = 0;
                while(j < 3)
                {
                    printf("Index %d in agument \"%s\" is %c\n", j, argv[i], argv[i][j]);
                    j++;
                }
        }
    }
    else
        fprintf(stderr,"Missing arguments\n");
        return(EXIT_FAILURE);

    return(EXIT_SUCCESS);
}

Answer 3

码：

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

const int num_chars = 3;
int main(int argc, char **argv) {
    if(argc == 1) return 1;
    if(num_chars <= 0) return 1;

    for(int i=1; i<argc; i++)
    {
       if(num_chars > strlen(argv[i])) return 1;
       printf("The first %d chars of %s is:\n", num_chars, argv[i]);
       for(int j=0; j<num_chars; j++)
       {
         printf("%d = %c\n", j, argv[i][j]);
       }
       printf("\n");
    }
    return 0;
}

有效情况：

$ ./test hello world
The first 3 chars of hello is:
0 = h
1 = e
2 = l

The first 3 chars of world is:
0 = w
1 = o
2 = r

无效的情况：

$ ./test hel wo
The first 3 chars of hel is:
0 = h
1 = e
2 = l