That
is my point, it doesn't fail. Either the kernel finds out a way to map the memory and it succeeds, or it kill the program and the instruction never runs. Code that doesn't run can't violate the standard. When the code is allowed to run all the invariants are guaranteed to be respected. If I write this code:
char *b = malloc(2);
if (b == NULL) {
return 0;
}
b[0] = 'A';
b[1] = '\0';
printf("%s\n", b);
free(b);
The standard tells me that if the malloc succeeds then the following code, if allowed to run, will display "A" on stdout. The C standard cannot and does not guarantee that a C program can't be interrupted however. For the sake of the argument we could imagine a kernel that instead of killing the program freezes it indefinitely on disk waiting for RAM to be available. It's functionally the same thing. As long as the kernel doesn't let code run with broken invariants it's fine. This is completely outside of the scope of a language standard to define.
Or, to try one last time from a different direction, if you consider that the C standard mandates that accessing memory returned successfully by malloc has to be successful and I happen to press ^C when that happens in a program, should the kernel refuse to kill the program? This is obviously absurd, but it's effectively the same thing: the kernel reacts to some external state and decides to terminate the program.