我有一个二进制文件，包括代表的ASCII文件整数。例如包含hello一个文件具有（使用XXD）的值
0000000: 4800 0000 6500 0000 6c00 0000 6c00 0000 H...e...l...l...
0000010: 6f00 0000 0000 0000 o.......
我怎样才能读取该文件，并将其转换为ASCII字符串Hello？
编辑的例子，显示的正是我做什么我得到。只需编译并运行它。将一个名为“test.txt的该PWD这就是里面的明文消息，结果是2个文件test.enc和test.dec。我知道它是不安全的所有，但我无论如何要创建此。
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#include <string.h>
#define ACCURACY 5
#define SINGLE_MAX 100000
#define EXPONENT_MAX 10000
#define BUF_SIZE 1024
#define DEBUG 0
int modpow(long long a, long long b, int c) {
int res = 1;
while(b > 0) {
/* Need long multiplication else this will overflow... */
if(b & 1) {
res = (res * a) % c;
}
b = b >> 1;
a = (a * a) % c; /* Same deal here */
}
return res;
}
int jacobi(int a, int n) {
int twos, temp;
int mult = 1;
while(a > 1 && a != n) {
a = a % n;
if(a <= 1 || a == n) break;
twos = 0;
while(a % 2 == 0 && ++twos) a /= 2; /* Factor out multiples of 2 */
if(twos > 0 && twos % 2 == 1) mult *= (n % 8 == 1 || n % 8 == 7) * 2 - 1;
if(a <= 1 || a == n) break;
if(n % 4 != 1 && a % 4 != 1) mult *= -1; /* Coefficient for flipping */
temp = a;
a = n;
n = temp;
}
if(a == 0) return 0;
else if(a == 1) return mult;
else return 0; /* a == n => gcd(a, n) != 1 */
}
int solovayPrime(int a, int n) {
int x = jacobi(a, n);
if(x == -1) x = n - 1;
return x != 0 && modpow(a, (n - 1)/2, n) == x;
}
int probablePrime(int n, int k) {
if(n == 2) return 1;
else if(n % 2 == 0 || n == 1) return 0;
while(k-- > 0) {
if(!solovayPrime(rand() % (n - 2) + 2, n)) return 0;
}
return 1;
}
int randPrime(int n) {
int prime = rand() % n;
n += n % 2; /* n needs to be even so modulo wrapping preserves oddness */
prime += 1 - prime % 2;
while(1) {
if(probablePrime(prime, ACCURACY)) return prime;
prime = (prime + 2) % n;
}
}
int gcd(int a, int b) {
int temp;
while(b != 0) {
temp = b;
b = a % b;
a = temp;
}
return a;
}
int randExponent(int phi, int n) {
int e = rand() % n;
while(1) {
if(gcd(e, phi) == 1) return e;
e = (e + 1) % n;
if(e <= 2) e = 3;
}
}
int inverse(int n, int modulus) {
int a = n, b = modulus;
int x = 0, y = 1, x0 = 1, y0 = 0, q, temp;
while(b != 0) {
q = a / b;
temp = a % b;
a = b;
b = temp;
temp = x; x = x0 - q * x; x0 = temp;
temp = y; y = y0 - q * y; y0 = temp;
}
if(x0 < 0) x0 += modulus;
return x0;
}
int readFile(FILE* fd, char** buffer, int bytes) {
int len = 0, cap = BUF_SIZE, r;
char buf[BUF_SIZE];
*buffer = malloc(BUF_SIZE * sizeof(char));
while((r = fread(buf, sizeof(char), BUF_SIZE, fd)) > 0) {
if(len + r >= cap) {
cap *= 2;
*buffer = realloc(*buffer, cap);
}
memcpy(&(*buffer)[len], buf, r);
len += r;
}
/* Pad the last block with zeros to signal end of cryptogram. An additional block is added if there is no room */
if(len + bytes - len % bytes > cap) *buffer = realloc(*buffer, len + bytes - len % bytes);
do {
(*buffer)[len] = '\0';
len++;
}
while(len % bytes != 0);
return len;
}
int encode(int m, int e, int n) {
return modpow(m, e, n);
}
int decode(int c, int d, int n) {
return modpow(c, d, n);
}
int* encodeMessage(int len, int bytes, char* message, int exponent, int modulus) {
int *encoded = malloc((len/bytes) * sizeof(int));
int x, i, j;
for(i = 0; i < len; i += bytes) {
x = 0;
for(j = 0; j < bytes; j++) x += message[i + j] * (1 << (7 * j));
encoded[i/bytes] = encode(x, exponent, modulus);
if(DEBUG) printf("%d ", encoded[i/bytes]);
}
return encoded;
}
int* decodeMessage(int len, int bytes, int* cryptogram, int exponent, int modulus) {
int *decoded = malloc(len * bytes * sizeof(int));
int x, i, j;
for(i = 0; i < len; i++) {
x = decode(cryptogram[i], exponent, modulus);
for(j = 0; j < bytes; j++) {
decoded[i*bytes + j] = (x >> (7 * j)) % 128;
if(DEBUG) if(decoded[i*bytes + j] != '\0') printf("%c", decoded[i*bytes + j]);
}
}
return decoded;
}
int main(void) {
int p, q, n, phi, e, d;
int len;
int bytes = 1;
int *encoded, *decoded;
char *buffer;
FILE *f;
srand(time(NULL));
while(1) {
p = randPrime(SINGLE_MAX);
q = randPrime(SINGLE_MAX);
n = p * q;
if(n < 128) {
printf("Modulus is less than 128, trying again\n");
} else {
break;
}
}
phi = (p - 1) * (q - 1);
e = randExponent(phi, EXPONENT_MAX);
d = inverse(e, phi);
// read the file
f = fopen("test.txt", "r");
if(f == NULL) return EXIT_FAILURE;
len = readFile(f, &buffer, bytes); // len will be a multiple of bytes, to send whole chunks
fclose(f);
if(DEBUG) printf("Encoded: ");
encoded = encodeMessage(len, bytes, buffer, e, n);
// save the encoded binary! to file
FILE *encodedFile;
encodedFile = fopen("tmp.enc", "wb");
fwrite(encoded, sizeof(int), len, encodedFile);
fclose(encodedFile);
// read the encodedFile binary! overwrite encoded!!!
FILE *newEncodedFile;
newEncodedFile = fopen("tmp.enc", "rb");
fread(encoded, sizeof(int), len, newEncodedFile);
fclose(newEncodedFile);
if(DEBUG) printf("\nDecoded: ");
decoded = decodeMessage(len/bytes, bytes, encoded, d, n);
// save the decoded binary! to file
FILE *decodedFile;
decodedFile = fopen("tmp.dec", "wb");
fwrite(decoded, sizeof(int), len/bytes, decodedFile);
fclose(decodedFile);
free(encoded);
free(decoded);
free(buffer);
return EXIT_SUCCESS;
}
字符是根据一些任意的规则显示只是整数。因为规则是任意有无限的可能性。那些无限可能性的存在多个标准（EBDIC，ASCII的“扩展ASCII”许多变化，统一等），因为他们都只是为整数，有许多方法来对其进行编码（“原样”，UTF-8，UTF -16LE，UTF-16BE，..）。
要的东西转换到任何东西，你需要解码原始编码（如果有的话），所产生的字符转换为新的字符集（可能通过。查找表），然后重新编码目标编码的字符（如果有的话） 。
你显示的数据看起来像“的Unicode与UTF-32LE编码”。为了将其转换为“ASCII与AS-被编码”（以便携的方式），你需要通过解码原始编码（例如codepoint = buffer[0] | (buffer[1] << 8) | (buffer[2] << 16) | (buffer[3] << 24)）开始。然后，你需要生成的代码点（“人物”）转换成ASCII字符集。幸运的是，第一码点128以Unicode是相同的ASCII。不幸的是，几乎所有的Unicode中的其他代码点不能被转换为ASCII，所以你需要决定如何处理这（在一个ASCII字符'?'取代他们？生成一个“不能转换”错误信息和放弃？）。在任何情况下，它可能会看起来像if(codepoint < 128) { character = codepoint; } else {。最后，由于ASCII采用了“原样”编码你可以一巴掌生成的字节到内存（有没有参与过“再编码”的工作）。
没有测试，但是这样的事情：
int utf32le2ascii(uint32_t *instr, size_t len, char *outstr, size_t outlen)
{
size_t i, j = 0;
for (i = 0; i < len; i++)
{
if (j >= outlen)
{
return -1;
}
if (instr[i] < 128)
{
outstr[j] = instr[i];
}
else
{
outstr[j] = '?';
}
j++;
}
if (j >= outlen)
{
return -1;
}
outstr[j] = '\0';
return 0;
}
-1的的返回值表示故障; 0表示成功。