我写这个函数是从文件中读取一行:
const char *readLine(FILE *file) {
if (file == NULL) {
printf("Error: file pointer is null.");
exit(1);
}
int maximumLineLength = 128;
char *lineBuffer = (char *)malloc(sizeof(char) * maximumLineLength);
if (lineBuffer == NULL) {
printf("Error allocating memory for line buffer.");
exit(1);
}
char ch = getc(file);
int count = 0;
while ((ch != '\n') && (ch != EOF)) {
if (count == maximumLineLength) {
maximumLineLength += 128;
lineBuffer = realloc(lineBuffer, maximumLineLength);
if (lineBuffer == NULL) {
printf("Error reallocating space for line buffer.");
exit(1);
}
}
lineBuffer[count] = ch;
count++;
ch = getc(file);
}
lineBuffer[count] = '\0';
char line[count + 1];
strncpy(line, lineBuffer, (count + 1));
free(lineBuffer);
const char *constLine = line;
return constLine;
}
该函数正确地读取文件，使用printf我看到constLine字符串也被正确读取。
然而，如果我像这样使用函数:
while (!feof(myFile)) {
const char *line = readLine(myFile);
printf("%s\n", line);
}
Printf输出胡言乱语。为什么?
在readLine函数中，返回一个指向行数组的指针(严格地说，是指向该行数组第一个字符的指针，但这里的区别无关紧要)。由于它是一个自动变量(即，它“在堆栈上”)，当函数返回时，内存将被回收。你看到的是乱七八糟的东西，因为printf把它自己的东西放到了堆栈上。
您需要从函数返回一个动态分配的缓冲区。你已经有一个了，它是lineBuffer;你所要做的就是把它截断成所需的长度。
lineBuffer[count] = '\0';
realloc(lineBuffer, count + 1);
return lineBuffer;
}
ADDED(对评论中后续问题的回应):readLine返回一个指向组成该行的字符的指针。这个指针是处理该行内容所需要的指针。当你用完这些字符占用的内存时，你也必须通过它来释放内存。下面是如何使用readLine函数:
char *line = readLine(file);
printf("LOG: read a line: %s\n", line);
if (strchr(line, 'a')) { puts("The line contains an a"); }
/* etc. */
free(line);
/* After this point, the memory allocated for the line has been reclaimed.
You can't use the value of `line` again (though you can assign a new value
to the `line` variable if you want). */
我想要一个代码从地面0，所以我这样做，逐行读取字典的单词的内容。
char temp_str [20];//你可以根据你的要求改变缓冲区的大小和一个文件中的单行长度。
注意，每次读取line时，我都将缓冲区初始化为Null字符。这个功能可以自动化，但因为我需要一个概念的证明，并想要一个字节一个字节地设计程序
#include<stdio.h>
int main()
{
int i;
char temp_ch;
FILE *fp=fopen("data.txt","r");
while(temp_ch!=EOF)
{
i=0;
char temp_str[20]={'\0','\0','\0','\0','\0','\0','\0','\0','\0','\0','\0','\0','\0','\0','\0','\0','\0','\0','\0'};
while(temp_ch!='\n')
{
temp_ch=fgetc(fp);
temp_str[i]=temp_ch;
i++;
}
if(temp_ch=='\n')
{
temp_ch=fgetc(fp);
temp_str[i]=temp_ch;
}
printf("%s",temp_str);
}
return 0;
}
实现从文件中读取和获取内容的方法(input1.txt)
#include <stdio.h>
#include <stdlib.h>
void testGetFile() {
// open file
FILE *fp = fopen("input1.txt", "r");
size_t len = 255;
// need malloc memory for line, if not, segmentation fault error will occurred.
char *line = malloc(sizeof(char) * len);
// check if file exist (and you can open it) or not
if (fp == NULL) {
printf("can open file input1.txt!");
return;
}
while(fgets(line, len, fp) != NULL) {
printf("%s\n", line);
}
free(line);
}
希望这对你有所帮助。编码快乐!
这个例子有一些错误:
you forgot to add \n to your printfs. Also error messages should go to stderr i.e. fprintf(stderr, ....
(not a biggy but) consider using fgetc() rather than getc(). getc() is a macro, fgetc() is a proper function
getc() returns an int so ch should be declared as an int. This is important since the comparison with EOF will be handled correctly. Some 8 bit character sets use 0xFF as a valid character (ISO-LATIN-1 would be an example) and EOF which is -1, will be 0xFF if assigned to a char.
There is a potential buffer overflow at the line
lineBuffer[count] = '\0';
If the line is exactly 128 characters long, count is 128 at the point that gets executed.
As others have pointed out, line is a locally declared array. You can't return a pointer to it.
strncpy(count + 1) will copy at most count + 1 characters but will terminate if it hits '\0' Because you set lineBuffer[count] to '\0' you know it will never get to count + 1. However, if it did, it would not put a terminating '\0' on, so you need to do it. You often see something like the following:
char buffer [BUFFER_SIZE];
strncpy(buffer, sourceString, BUFFER_SIZE - 1);
buffer[BUFFER_SIZE - 1] = '\0';
if you malloc() a line to return (in place of your local char array), your return type should be char* - drop the const.
