问题描述
我有一个名为的文本文件的test.txt
I have a text file named test.txt
我想写一个C程序可以读取该文件与打印内容到控制台(假设该文件只包含ASCII文本)。
I want to write a C program that can read this file and print the content to the console (assume the file contains only ASCII text).
我不知道如何让我的字符串变量的大小。像这样的:
I don't know how to get the size of my string variable. Like this:
char str[999];
FILE * file;
file = fopen( "test.txt" , "r");
if (file) {
while (fscanf(file, "%s", str)!=EOF)
printf("%s",str);
fclose(file);
}
尺寸 999
不起作用,因为的fscanf
返回的字符串可以比大。我怎样才能解决这个问题?
The size 999
doesn't work because the string returned by fscanf
can be larger than that. How can I solve this?
推荐答案
最简单的方法是读取一个字符,阅读后可即时列印:
The simplest way is to read a character, and print it right after reading:
int c;
FILE *file;
file = fopen("test.txt", "r");
if (file) {
while ((c = getc(file)) != EOF)
putchar(c);
fclose(file);
}
C
是 INT
上面,因为 EOF
是一个负数,而一个普通的字符
可能是无符号
。
c
is int
above, since EOF
is a negative number, and a plain char
may be unsigned
.
如果你想阅读数据块中的文件,但没有动态内存分配,可以这样做:
If you want to read the file in chunks, but without dynamic memory allocation, you can do:
#define CHUNK 1024 /* read 1024 bytes at a time */
char buf[CHUNK];
FILE *file;
size_t nread;
file = fopen("test.txt", "r");
if (file) {
while ((nread = fread(buf, 1, sizeof buf, file)) > 0)
fwrite(buf, 1, nread, stdout);
if (ferror(file)) {
/* deal with error */
}
fclose(file);
}
上述的第二种方法是基本上将如何读取具有动态分配阵列的文件:
The second method above is essentially how you will read a file with a dynamically allocated array:
char *buf = malloc(chunk);
if (buf == NULL) {
/* deal with malloc() failure */
}
/* otherwise do this. Note 'chunk' instead of 'sizeof buf' */
while ((nread = fread(buf, 1, chunk, file)) > 0) {
/* as above */
}
您的方法的fscanf()
与%S
为格式文件中损失大约空白信息,因此它不正是将文件复制到标准输出
。
Your method of fscanf()
with %s
as format loses information about whitespace in the file, so it is not exactly copying a file to stdout
.
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