Code 1:
unsigned int *p = malloc(sizeof *p);
memset(p, 0x55, sizeof *p);
unsigned int u = *p;
Code 2:
void *d = malloc(50);
*(double *)d = 1.23;
memset(d, 0x55, 50);
unsigned int u = *(unsigned int *)d;
In each case, what effect does memset
have on the effective type of the object in the malloc'd space; and so is initializing u
correct or a strict aliasing violation?
The definition of effective type (C11 6.5/6) is:
The effective type of an object for an access to its stored value is the declared type of the object, if any. If a value is stored into an object having no declared type through an lvalue having a type that is not a character type, then the type of the lvalue becomes the effective type of the object for that access and for subsequent accesses that do not modify the stored value. If a value is copied into an object having no declared type using memcpy
or memmove
, or is copied as an array of character type, then the effective type of the modified object for that access and for subsequent accesses that do not modify the value is the effective type of the object from which the value is copied, if it has one. For all other accesses to an object having no declared type, the effective type of the object is simply the type of the lvalue used for the access.
However it is unclear whether memset
behaves like writing through an lvalue of character type, or something else. The description of memset
(7.24.6.1) is not very illuminating:
The memset function copies the value of c (converted to an unsigned char
) into each of the first n characters of the object pointed to by s.
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