char const c = 'x'; /* 1 */
char *p1; /* 2 */
char const **p2 = &p1; /* 3 */
*p2 = &c; /* 4 */
*p1 = 'X'; /* 5 */
Line 3 is disallowed in C and is invalid in C++. However, one gets an error in C++, but only a warning in C. The question is - why only a warning?
Both C & C++ deal with statement 3 as that "Assigning const char**
with char**
discards the const
qualifier" The difference is; C++ is more strict than C and doesn't allow const
qualifier discarding. This is really convenient in C++ and does lead to no issues contrary to C.
Imagine the following code:
char const c = 'x';
char *p1 = &c;
*p1 = 'A';
printf("c address: 0x%X, p1 is pointing to: 0x%X\n", &c, p1);
printf("%c\n", c);
printf("%c\n", *p1);
This is a compilable C code. One would think that assigning a new value to *p1
should not be possible as it points to a const
variable. Also one would think that if it did happen and we could change the value of what p is pointing at this will also change the value of c
But this is not the case!!
The above code surprisingly generates:
c address: 0xEF3E146B, p1 is pointing to: 0xEF3E146B
x
A
So allowing this const qualifier to be discarded does lead to undefined behavior (which means it's up to the compiler to decide what to do). This is fixed in C++ to avoid such ambiguities.
C has much more relaxed pointer rules than C++. C++ enforces a lot more type safety than C. Some of this is up to compilers to decide only to warn in C.
One example to look at is malloc, where char* str = malloc(50);
is perfectly fine C but must be char* str = (char*) malloc(50);
in C++.
The short answer is: because they are different languages and have different rules. There is no C/C++ language.
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