14. warning passing argument 1 of 'fopen' from incompatible pointer type
19. warning passing argument 1 of 'fopen' from incompatible pointer type

fopen() 我怎么看都没用错啊。

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

int main(int argc,int *argv[])
{
    FILE *fin,*fout;
    char a[1000];

    if(argc != 3)
    {
        printf("参数错误.\n");
        exit(EXIT_FAILURE);
    }
    if((fin = fopen(argv[1],"rb")) == NULL)
    {
        printf("无法打开源文件.\n");
        exit(EXIT_FAILURE);
    }
    if((fout = fopen(argv[2],"wb")) == NULL)
    {
        printf("无法打开目标文件.\n");
        exit(EXIT_FAILURE);
    }
    while((fread(a,sizeof(char),1000,fin)) > 0)
        fwrite(a,sizeof(char),1000,fout)
    return 0;
}

好吧………………原来是这里，看来下次不该老盯着提示的行数看。