Is this valid C code without undefined behaviour?
int main(){
int a;
memset(&a, 5, sizeof(int));
return a;
}
I'm assuming this is equal to just doing int a = 5.
I'm trying to understand if just declaring a variable in the above example (without defining it) is enough to put it on the stack.
Is this valid C code without undefined behaviour?
Yes – Once the a variable has been declared in a given scope (like a function or other { ... } delimited block), it is valid to take its address and access the variable using that address within that scope (as your memset call does). An attempt to use that address when that scope has ended (i.e. is no longer 'active') will cause undefined behaviour; for example, the following is UB:
int main()
{
int* p;
{ // New scope ...
int a;
p = &a; // Pointer to "a" is valid HERE
} // The scope of "a" (and its 'lifetime') ends here
memset(p, 5, sizeof(int)); // INVALID: "p" now points to a dead (invalid) variable
}
However, there's a major caveat in your code sample …
I'm assuming this is equal to just doing int a = 5.
There's the rub: It's assigning 5 to each component byte of the a variable, so it's doing this (assuming a 4-byte int):
int a = 0x05050505;
Which is the same as:
int a = 84215045;
char.
{ ...} delimited block), taking its address (and using it, mostly) is valid.From the C Standard (7.23.6.1 The memset function)
2 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.
So this call
memset(&a, 5, sizeof(int));
does not set the variable a equal to 5. Internally the variable will look like
0x05050505
Here is a demonstrative program
#include <stdio.h>
#include <string.h>
int main(void)
{
int a;
memset( &a, 5, sizeof( int ) );
printf( "%#x\n", ( unsigned )a );
return 0;
}
Its output is
0x5050505
You should use the function memset with integers with caution because in general it can produce a trap value. Also the result depends on how internally integers are stored starting from MSB or LSB.
P.S. You declared a variable inside a block scope with no linkage. It is also a variable definition that has automatic storage duration. As the variable explicitly was not initialized then it has an indeterminate value. You may apply the address of operator & to get the address of the memory extent where the variable is defined.
char.That's not undefined behavior. The problem is that it doesn't what you expect.
The result of
memset(&a, 5, sizeof(int));
consists in setting to 5 each of the four bytes of your integer a.
int a = 5.