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Why can’t I convert ‘char**’ to a ‘const char* const*’ in C?
我很好奇，为什么我不能将 char ** 传递给 const char ** 函数？虽然可以将 char * 传递给 const char * 函数，但使用双指针似乎不行。我认为添加 constness 总是可以的(但不可以删除 constness)但现在看来我错了。
Gcc 编译器给我错误:
note: expected ‘const char **’ but argument is of type ‘char **’
这是代码片段:
int f(const char **a) { }
int main() {
char *a;
f(&a);
}
有什么想法吗？
最佳答案
因为编译器不能保证安全。
请参阅 comp.lang.c FAQ 中的 Q11.10:Why can't I pass a char ** to a function which expects a const char **?
suppose you performed the following more complicated series of assignments:
const char c = 'x'; /* 1 */ char *p1; /* 2 */ const char **p2 = &p1; /* 3 */ *p2 = &c; /* 4 */ *p1 = 'X'; /* 5 */In line 3, we assign a
char **to aconst char **. (The compiler should complain.) In line 4, we assign aconst char *to aconst char *; this is clearly legal. In line 5, we modify what achar *points to--this is supposed to be legal. However,p1ends up pointing toc, which isconst. This came about in line 4, because*p2was reallyp1. This was set up in line 3, which is an assignment of a form that is disallowed, and this is exactly why line 3 is disallowed.Assigning a
char **to aconst char **(as in line 3, and in the original question) is not immediately dangerous. But it sets up a situation in whichp2's promise--that the ultimately-pointed-to value won't be modified--cannot be kept.
关于c - 为什么不能将 `char **` 传递给在 C 中采用 `const char **` 的函数？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/3496000/