编写一个程序,从中接收三个参数 命令行:week09_1 file1.txt file2.txt file3.txt
您的程序应该打开文件file1.txt和file2.txt进行阅读并创建
file file3.txt文件如下:
file3.txt的第一行是file1.txt的第一行 file3.txt的第二行是file2.txt的第一行 file3.txt的第三行是file1.txt的第二行 file3.txt的第四行是file2.txt的第二行
...
当一个输入文件到达EOF时,另一个文件中的剩余行为
应该复制到输出文件,程序终止。您
如果少于3个文件名,程序应打印相应的错误消息
在命令行上提供或无法打开文件。
截至目前,我已经能够做出提示要求我做的事情;但是,当一个文件用完要打印的行,然后它只从另一个文件打印时,打印在同一行,我想知道如何从1个文件打印时才能启动新行。
此外,我不明白我是如何在命令行参数中以提示的方式实现的。
#include <stdio.h>
char line1[256] = { };
char line2[256] = { };
char check;
int END = 0, END1 = 0, END2 = 0;
int main(int argc, char *argv[]) {
if (argc != 4) {
printf("Error: Wrong amount of arguments\n");
return 0;
}
FILE *file1 = fopen("argv[1]", "r");
FILE *file2 = fopen("argv[2]", "r");
FILE *file3 = fopen("argv[3]", "w");
if (argv == NULL) {
printf("Error: file could not be opened.\n");
return 0;
}
while (END != 2) {
check = fgets(line1, 256, file1);
if (check != NULL) {
fprintf(file3, "%s", line1);
} else if (check == NULL) {
END1 = 1;
}
check = fgets(line2, 256, file2);
if (check != NULL) {
fprintf(file3, "%s", line2);
} else if (check == NULL) {
END2 = 1;
}
END = END1 + END2;
}
return 0;
}
答案 0 :(得分:1)
我会让你在那里的一部分。
将此文件指定为file1.txt
:
File 1, Line 1
File 1, Line 2
File 1, Line 3
File 1, Line 4
File 1, Line 5
File 1, Line 6
File 1, Line 7
File 1, Line 8
此文件为file2.txt
:
File 2, Line 1
File 2, Line 2
File 2, Line 3
File 2, Line 4
File 2, Line 5
File 2, Line 6
File 2, Line 7
File 2, Line 8
File 2, Line 9
File 2, Line 10
您可以阅读它们并将其打印出来,如下所示:
#include <stdio.h>
int main(int argc, char *argv[]){
char buffer[4096];
char *ptr;
int i=0;
FILE *file1 = fopen("/tmp/file1.txt", "r");
FILE *file2 = fopen("/tmp/file2.txt", "r");
// while there is something from either file, do...
while (!feof(file1) || !feof(file2)){
// use modulo to switch file1, file2, etc
if (i%2 == 0) {
ptr=fgets(buffer, sizeof(buffer), file1);
}
else {
ptr=fgets(buffer, sizeof(buffer), file2);
}
// test if the last fgets actually read something by testing ptr vs NULL
if (ptr)
printf("%s", buffer);
++i;
}
return 0;
}
打印:
File 1, Line 1
File 2, Line 1
File 1, Line 2
File 2, Line 2
File 1, Line 3
File 2, Line 3
File 1, Line 4
File 2, Line 4
File 1, Line 5
File 2, Line 5
File 1, Line 6
File 2, Line 6
File 1, Line 7
File 2, Line 7
File 1, Line 8
File 2, Line 8
File 2, Line 9
File 2, Line 10
您需要将打印添加到文件3,文件上适当的fclose
,读取argv以及相应的错误测试/反应。
答案 1 :(得分:1)
我认为代码更容易管理,特别是如果文件中的最后一行没有换行符,如果使用函数处理文件的一行,并重复调用它。
#include <assert.h>
#include <stdio.h>
#include <string.h>
static int read_and_print_line(FILE *ifp, FILE *ofp)
{
char buffer[64];
int eol_needed = 0;
while (fgets(buffer, sizeof(buffer), ifp) != 0)
{
fputs(buffer, ofp);
if (strchr(buffer, '\n') != 0)
return 0;
/* Either more to read in current line or EOF without newline */
eol_needed = 1;
}
if (eol_needed)
putc('\n', ofp);
return EOF;
}
int main(int argc, char **argv)
{
if (argc != 4)
{
fprintf(stderr, "Usage: %s in-file1 in-file2 outfile\n", argv[0]);
return 1;
}
FILE *if1 = fopen(argv[1], "r");
FILE *if2 = fopen(argv[2], "r");
FILE *ofp = fopen(argv[3], "w");
if (if1 == 0)
fprintf(stderr, "Failed to open %s for reading\n", argv[1]);
if (if2 == 0)
fprintf(stderr, "Failed to open %s for reading\n", argv[2]);
if (ofp == 0)
fprintf(stderr, "Failed to open %s for reading\n", argv[3]);
if (if1 == 0 || if2 == 0 || ofp == 0)
return 1;
int r1 = 0;
int r2 = 0;
while ((r1 = read_and_print_line(if1, ofp)) != EOF &&
(r2 = read_and_print_line(if2, ofp)) != EOF)
;
assert(r1 == EOF || r2 == EOF);
assert(r1 != r2);
while (read_and_print_line(if1, ofp) != EOF)
;
while (read_and_print_line(if2, ofp) != EOF)
;
fclose(if1);
fclose(if2);
fclose(ofp);
return 0;
}
我很满意这可以在它自己的源代码上工作,也可以在一个'行'文件上工作,最后没有换行符(echo -n Gobbledygook >file1
),依此类推。缓冲区显示为64字节以帮助测试(如果您愿意,可以将其减少到32或甚至更低)。经过测试,我的编译大小为4096。
答案 2 :(得分:0)
一些UB代码需要修复
//char check;
void * check;
推荐:
char line1[256] = { 0 };
char line2[256] = { 0 };
更改错误检查。可以扩展哪个文件不好
if (file1 == NULL || file2 == NULL || file3 == NULL) {
printf("Error: file could not be opened.\n");
return 0;
}
但其他人有一个很好的“其余答案”。