使用mytee合并makefile时如何解决错误

如何解决使用mytee合并makefile时如何解决错误

当我使用getopt API完成代码（mytee.c）以便在UNIX中编写实现命令并使用make test测试文件时，我得到以下消息：

"make test
gcc -o mytee mytee.c
make test1
make[1]: Entering directory '/home/programmer/proj1'
./mytee -v
usage: mytee [-av] _filename_
make[1]: Leaving directory '/home/programmer/proj1'
make test2
make[1]: Entering directory '/home/programmer//proj1'
cat mytee.c | ./mytee mytee.fout > mytee.out
Expected argument after options
make[1]: *** [Makefile:32: test2] Error 1
make[1]: Leaving directory '/home/programmer/proj1'
make: *** [Makefile:21: test] Error 2" 

运行测试时，将创建以下文件：mytee.fout mytee.out mytee random.bin

makefile是这样的： [makefile] [1]

当我执行此命令diff mytee.fout mytee.c时，收到以下消息：0a1,77。在此处输入图片说明。为了正确运行，我应该怎么做才能解决此错误？ 我的代码在这里：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<unistd.h>
#include<assert.h>


#define USAGE_MESSAGE "usage: mytee [-av] _filename_"

/* globals and externs go here */


int main(int argc,char * argv[]) {
    int ch ;
    int is_verbose = 0 ;
    int is_append = 0 ;
    char * mode_s = NULL ;
    char * filename ;
    FILE * f ;
    int nsecs =0;
    while ((ch = getopt(argc,argv,"va")) != -1) {
        switch(ch) {
        /*
         * modify or add to these case statements
         */
        case 'v':
            is_verbose = 1 ;
            break ;
        case 'a':
            mode_s= optarg;
            is_append = 1 ;
            break ;
        default:
                printf("%s\n",USAGE_MESSAGE) ;
                return 0 ;
        }
    }
    argc -= optind; 
    argv += optind;

    if (argc!=1) {
            printf("%s\n",USAGE_MESSAGE) ;
        return 0 ;
    }

    filename = argv[0] ;
    if (is_verbose) { 
        printf("%s:%d: openning file %s in mode %s\n",__FILE__,__LINE__,filename,mode_s) ;
    }
    f = fopen(filename,mode_s) ;
    assert(f) ;
    while (1) {
    
    while (ch = getchar()&&ch != EOF) {
        putchar(ch);
        ch = getchar();
    }
    printf("verbose=%d; append=%d; optind=%d\n",is_verbose,is_append,optind);
      printf("name argument = %s\n",argv[optind]);
    break;
    }
     if (optind >= argc) {
        fprintf(stderr,"Expected argument after options\n");
        exit(EXIT_FAILURE);
    }
     return 0; }

makefile就在这里

make mytee

test: mytee
        make test1
        make test2
        make clean
        @echo "*** ${GREEN}PASSED the test ${NC}***"

mytee: mytee.c
        gcc -o mytee mytee.c

test1: mytee
        ./mytee -v

test2: mytee
        cat mytee.c | ./mytee mytee.fout > mytee.out
        diff mytee.fout mytee.c
        diff mytee.c mytee.out

test3: mytee
        dd if=/dev/urandom bs=1 count=4096 of=random.bin
        cat random.bin | ./mytee mytee.fout > mytee.out
        diff random.bin mytee.fout
        diff random.bin mytee.out


mytest: mytee
        @echo "student adds tests,for instance,"
        @echo "* mytee called without argument,"
        @echo "* mytee called with -a option"

clean:
        -rm mytee.fout mytee.out mytee random.bin

解决方法

参数列表包括程序名称。该程序的第一个参数在argv[1]中。

所以如果你打电话

./mytee

你会得到

argc:    1
argv[0]: "./mytee"

产生错误的命令是

./mytee mytee.fout

您将得到

argc:    2
argv[0]: "./mytee"
argv[1]: "mytee.fout"

但是您正在将argc与1（而非2）进行比较。

---

注意：您可以通过阅读文档来找到这些东西，或者通过编写像这样的简单测试来发现这些东西：

#include <stdio.h>

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

    for (i = 0; i < argc; ++i) {
        printf("%d: \"%s\"\n",i,argv[i]);
    }
}