我有以下两个功能:
void compute_key(uint8_t * stream, int key_length, int stream_length, uint8_t * key) {
// there will be key_length encrypted streams;
uint8_t ** encr_streams;
int * bytes_in_stream;
encr_streams = (uint8_t **)malloc(key_length * sizeof **encr_streams);
bytes_in_stream = (int *)malloc(key_length * sizeof *bytes_in_stream);
for (int i = 0; i < key_length; i++) bytes_in_stream[i] = 0;
construct_cypherstreams(stream, key_length, stream_length, encr_streams, bytes_in_stream);
printf("%s\n", "bytes_in_stream[]");
for (int i = 0; i < key_length; i++) {
printf("%d\n", bytes_in_stream[i]);
}
getchar();
for (int i = 0; i < key_length; i++) bytes_in_stream[i] = 7; // introduced in step #2
printf("\n%s\n\n", "cypherstreams:");
for (int i = 0; i < key_length; i++) {
for (int j = 0; j < bytes_in_stream[i]; j++) {
printf("%X", encr_streams[i][j]);
}
printf("\n\n");
}
// for each cypherstream, compute ByteB - the value they were XOR`d with
compute_key_using_scoring(encr_streams, bytes_in_stream, key_length, key);
getchar();
}
void construct_cypherstreams(uint8_t * stream, int key_length, int stream_length, uint8_t ** encr_streams, int * bytes_in_stream) {
// chyperstream = the stream formed of every ith byte
uint8_t * cypherstream;
int length;
length = stream_length / key_length + 1;
// each byte of the key can have values
// between 0 and 256
int i = 0;
int num_added = 0;
for (int k = 0; k < key_length; k++) {
i = k; num_added = 0;
cypherstream = (uint8_t *)malloc(length * sizeof *cypherstream);
if (cypherstream == NULL) {
printf("%s\n", "could not allocate");
exit(1);
}
while (i < stream_length) {
// construct cypherstream
cypherstream[num_added] = stream[i];
num_added++;
i += key_length;
}
// this is always correct
printf("\n%s\n", "created cypherstream:");
for (int m = 0; m < num_added; m++) {
printf("%X", cypherstream[m]);
}
printf("\n");
encr_streams[k] = cypherstream;
bytes_in_stream[k] = num_added;
// this is always correct
printf("%s%d%s %d\n", "bytes_in_stream[", k, "]", bytes_in_stream[k]);
}
}
我的key_length
是31
,我确信它很好(之前计算过,也适用于其他值,对于这些值,程序到达末尾并打印解密的 chypertexht)。
在construct_cypherstreams
中,我将指针传递给compute_key
中动态分配的内存。
我的目的是在编译时填充未知大小的 2D 数组 encr_streams
和 1D 数组 bytes_in_stream
。
在 construct_cypherstreams
中,我在每次迭代后打印 bytes_in_stream[k]
的内容,并获得有效数据。每个流有 7 个字节,一切都按预期工作。我还打印了新形成的 encr_streams[k]
并且也得到了预期值。
在第 1 步,程序从 construct_cypherstreams
返回后,我检查 bytes_in_stream
的内容是否正确,并且我看到前 7 个值是垃圾我真的不明白为什么。我已经编写了很多虚拟示例来填充函数内的各种数组,它们似乎都有效。
在第 2 步,当我感到非常沮丧时,我只是用我知道它应该包含的值填充 bytes_in_stream
,以便我可以进入代码的下一部分。令我惊讶的是,当我尝试打印 encr_streams
时,当 i
变为 24
(总共 31 个 encr_streams
)。
我真的不明白为什么会发生这种情况,我感觉我遇到了未定义行为的情况。
当发现另一个流的 key_length
为 6
时,程序运行顺利。
最佳答案
这一行是错误的:
encr_streams = (uint8_t **)malloc(key_length * sizeof **encr_streams);
应该是:
encr_streams = (uint8_t **)malloc(key_length * sizeof *encr_streams);
关于c - 为什么内存会损坏?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27187818/