英文:
Understand Strlen applied to int array with char * cast
问题
I'm currently stuck on this problem.
I have the following code:
int v[10] = {-1, 1000, 2};
int a;
a = strlen((char *)v+1);
printf("strlen result for (char *) v+1: %d\n", a);
I cannot understand why the final result is always 5 on a little endian "machine".
According to strlen documentation, the char * cast in this situation should return 4.
I tried also in an online compiler and other machines, but the result remains the same. What am I missing?
英文:
I'm currently stuck on this problem.
I have thefollowing code:
int v[10] = {-1, 1000, 2};
int a;
a = strlen((char *)v+1);
printf("strlen result for (char *) v+1: %d\n", a);
I cannot undestand why the final result is always 5 on a little endian "machine".
According to strlen documentation, the char * cast in this situation should return 4.
I tried also in online compiler and other machines but the result remains the same. What am I missing ?
答案1
得分: 5
最可能的是 V 中的字节表示为：
0xff, 0xff, 0xff, 0xff, 0xe8, 0x03, 0x00, 0x00, ...
所以在跳过第一个字节之后（而不是第一个整数，因为你将其转换为 char *），在第一个 '\0' 字符之前有 5 个字节。
英文:
Most likely the bytes in V are represented as:
0xff, 0xff, 0xff, 0xff, 0xe8, 0x03, 0x00, 0x00, ...
So after skipping the first byte (not the first int, since you cast to char *) there are 5 bytes before the first '\0' character.
答案2
得分: 4
不应在整数数组上使用 strlen。函数的名称已经表明它只应用于合法形式的C字符串。
也就是说，出于学习的目的，对于小端序机器上的数组 v，其内存布局看起来像这样：
ff ff ff ff e8 03 00 00 02 00 00 00 (00 向后)
低 0>-^^----<0 1>-------<1 2>-------<2 3> ...----<9 高
由于你在调用 strlen((char *)v + 1)，计算从我标记为 ^ 的位置开始。有5个非零元素 (ff, ff, ff, e8, 03)，所以结果是5，而不是4。请记住 1000 等于 0x3e8。
英文:
You should not use strlen on an int array. The function's name already suggested it should only be used on legally formed C strings.
That being said, for the reason of studying, array v's memory layout looks something like this on a little-endian machine:
ff ff ff ff e8 03 00 00 02 00 00 00 (00 onwards)
low 0>-^^----<0 1>-------<1 2>-------<2 3> ...----<9 high
Since you are calling strlen((char *)v + 1), the computation starts at the location I marked with ^. There're 5 non-zero elements (ff, ff, ff, e8, 03), so the result is 5, not 4. Remember 1000 == 0x3e8.
答案3
得分: 3
这个小演示可以帮助你澄清小端机器中字节的顺序（假设一切都按上文所写）：
char* z = (char*)v + 1;
for (int i = 0; i < 10; i++)
printf("%02x ", (uint8_t)z[i]);
输出应该像这样：
ff ff ff e8 03 00 00 02 00 00（就像Harlan Wei上面展示的一样）。
这个小片段还可以帮助你看到在大端机器中字节是如何存储的。因此，简单的计数应该让你得到5，而不是4，就像strlen(...)试图说明的那样。
英文:
This little demo can help you clarify how bytes are ordered in little-endian machine (presuming every thing is as written above):
char* z = (char*)v + 1;
for (int i = 0; i < 10; i++)
printf("%02x ", (uint8_t)z[i]);
the print out should be like this:
ff ff ff e8 03 00 00 02 00 00 (like Harlan Wei showed above).
this little snippet can also help you see how bytes are stored in big-endian machines too. So mere counting should give you 5 not 4 as strlen(...) was trying to say.
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