C question

is a byte with all zeros equivalent to the integer value (signed or unsigned) zero, in terms of portability?

Such that, is memset(&i, 0, sizeof(int)); equivalent to i = 0;?

>>1 why not look at the source code
https://github.com/lattera/glibc/blob/master/sysdeps/i386/memset.c#L30
memset (void *dstpp, int c, size_t len)
{
int d0;
unsigned long int dstp = (unsigned long int) dstpp;

/* This explicit register allocation
improves code very much indeed. */
register op_t x asm("ax");

x = (unsigned char) c;

/* Clear the direction flag, so filling will move forward. */
asm volatile("cld");

/* This threshold value is optimal. */
if (len >= 12)
{
/* Fill X with four copies of the char we want to fill with. */
x |= (x << 8);
x |= (x << 16);

/* Adjust LEN for the bytes handled in the first loop. */
len -= (-dstp) % OPSIZ;

/* There are at least some bytes to set.
No need to test for LEN == 0 in this alignment loop. */

/* Fill bytes until DSTP is aligned on a longword boundary. */
asm volatile("rep\n"
"stosb" /* %0, %2, %3 */ :
"=D" (dstp), "=c" (d0) :
"0" (dstp), "1" ((-dstp) % OPSIZ), "a" (x) :
"memory");

/* Fill longwords. */
asm volatile("rep\n"
"stosl" /* %0, %2, %3 */ :
"=D" (dstp), "=c" (d0) :
"0" (dstp), "1" (len / OPSIZ), "a" (x) :
"memory");
len %= OPSIZ;
}

/* Write the last few bytes. */
asm volatile("rep\n"
"stosb" /* %0, %2, %3 */ :
"=D" (dstp), "=c" (d0) :
"0" (dstp), "1" (len), "a" (x) :
"memory");

return dstpp;
}
libc_hidden_builtin_def (memset)