本人写了一个
#include "StdAfx.h"
#include <stdio.h>
#include <stdlib.h>
//using namespace std;
struct BlockInfo {
double x;
double y;
double h;
};
int main() {
//FILE *f1=fopen("c:\\1\\2.csv","r");
FILE* f = fopen("c:\\1\\booltest", "w");
BlockInfo* blockInfo = (BlockInfo*)malloc(sizeof(blockInfo));
//493161.66,4478847.05,188
blockInfo->x=493161.66;/////将该值改为 193161.66 就占 24个字节 可没有溢出呀？ 问题载哪呢？？？
blockInfo->y=4478847.05;
blockInfo->h=188.00;
fwrite(blockInfo, sizeof(BlockInfo), 1, f);
fclose(f);
printf("ok");
return 0;
}
将数据写到二进制文件的程序，三个双精度数，本以为占24个字节，实际是25个，求助大神！

有任何问题可以进来大佬们免费解答

3D 0D 0A D7 A3 A6 19 1E 41 33 33 33 C3 DF 15 51
41 00 00 00 00 00 80 67 40
25字节
7B 14 AE 47 4D 94 07 41 33 33 33 C3 DF 15 51 41
00 00 00 00 00 80 67 40
24字节
从文件读出
int main() {
//FILE *f1=fopen("c:\\1\\2.csv","r");
FILE* f = fopen("booltest", "w");
BlockInfo* blockInfo = (BlockInfo*)malloc(sizeof(blockInfo));
//493161.66,4478847.05,188
//blockInfo->x=493161.66;/////将该值改为 193161.66 就占 24个字节 可没有溢出呀？ 问题载哪呢？？？
blockInfo->x=193161.66;
blockInfo->y=4478847.05;
blockInfo->h=188.00;
printf("x=%f\n",blockInfo->x);
printf("y=%f\n",blockInfo->y);
printf("z=%f\n",blockInfo->h);
printf("size=%d\n",sizeof(BlockInfo));
fwrite(blockInfo, sizeof(BlockInfo), 1, f);
fclose(f);
printf("ok\n\n");
f = fopen("booltest", "r");
fread(blockInfo, sizeof(BlockInfo), 1, f);
printf("x=%f\n",blockInfo->x);
printf("y=%f\n",blockInfo->y);
printf("z=%f\n",blockInfo->h);
fclose(f);
system("pause");
return 0;
}
x=193161.660000
y=4478847.050000
z=188.000000
size=24
ok
x=193161.660000
y=4478847.050000
z=188.000000
请按任意键继续. . .
读出没错
x=493161.660000
y=4478847.050000
z=188.000000
size=24
ok
x=493161.660000
y=4478847.050000
z=188.000000
请按任意键继续. . .
读出也没错