With C99 it's not that obvious. Compound literals are outside of C++ standard. However GCC for example supports them, but a bit differently [0].

  $ cat foobar.c
  #include <stdio.h>
  
  struct foo {
  	int x;
  	int y;
  };
  
  int foobar(struct foo *f)
  {
  	return f->x + f->y;
  }
  
  int
  main(int argc, char *argv[]) {
  	(void)argc; (void)argv;
  
  	printf("%d\n", foobar(&(struct foo){10, 20}));
  	return 0;
  }
  $ gcc -Wall -Wcast-align -Wextra -pedantic -std=c99 foobar.c -o foobar && ./foobar
  30
  $ g++ -Wall -Wcast-align -Wextra -pedantic foobar.c -o foobar && ./foobar
  foobar.c: In function ‘int main(int, char**)’:
  foobar.c:17: warning: ISO C++ forbids compound-literals
  foobar.c:17: warning: taking address of temporary
  30
  $ # taking address of temporary means that it's undefined behavior

[0] https://gcc.gnu.org/onlinedocs/gcc/Compound-Literals.html

The stated he was only using the subset of C99 that could compile as C++.