为什么fwrite的写作比我说的要多？

Question

FILE *out=fopen64("text.txt","w+");
unsigned int write;
char *outbuf=new char[write];
//fill outbuf
printf("%i\n",ftello64(out));
fwrite(outbuf,sizeof(char),write,out);
printf("%i\n",write);
printf("%i\n",ftello64(out));

输出:

0
25755
25868

到底是怎么回事？write设置为25755,我告诉fwrite将多个字节写入文件,这是在开头,然后我在25755以外的位置？

Answer 1

如果您使用的是DOSish系统(例如,Windows)并且文件未以二进制模式打开,则行结尾将自动转换,每个"行"将添加一个字节.

因此,指定"wb"为模式而不是"w"像@caf指出的那样.它对Unix平台没有任何影响,并且会对其他人做正确的事情.

例如:

#include <stdio.h>

#define LF 0x0a

int main(void) {
    char x[] = { LF, LF };

    FILE *out = fopen("test", "w");

    printf("%d", ftell(out));
    fwrite(x, 1, sizeof(x), out);
    printf("%d", ftell(out));

    fclose(out);
    return 0;
}

使用VC++:

C:\Temp> cl y.c
Microsoft (R) 32-bit C/C++ Optimizing Compiler Version 15.00.21022.08 for 80x86
Copyright (C) Microsoft Corporation.  All rights reserved.

y.c
Microsoft (R) Incremental Linker Version 9.00.21022.08
Copyright (C) Microsoft Corporation.  All rights reserved.

/out:y.exe

C:\Temp> y.exe
04

使用Cygwin gcc:

/cygdrive/c/Temp $ gcc y.c -o y.exe

/cygdrive/c/Temp $ ./y.exe
02