数据在内存中的存储形式

出题脚本

#C语言
#include <stdio.h>
char flag[]="afctf{sec_is_everywhere}";

int main()
{
	for(int i=0;i<6;++i){
		printf("%20f\n",*(float*)(flag+i*4));
	}
	return 0;
}

解题脚本

from libnum import*
import struct

s = [72065910510177138000000000000000.000000,71863209670811371000000.000000,18489682625412760000000000000000.000000,72723257588050687000000.000000,4674659167469766200000000.000000,19061698837499292000000000000000000000.000000]
a = ''
b = ''
for i in s:
    i = float(i)
    a += struct.pack('<f',i).hex()        #小端
print(a)

for j in s:
    i = float(i)
    b += struct.pack('>f',i).hex()        #大端
print(b)
a = 0x61666374667b7365635f69735f657665727977686572657d
b = 0x7d6572657d6572657d6572657d6572657d6572657d657265
print(n2s(a))
print(n2s(b))