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In the following noncompliant code example, if malloc() is not declared, either explicitly, or by including stdlib.h, a compiler may implicitly declare malloc() as int malloc(). (Compilers that only comply with C90 are required to provide an implicit declaration of malloc().)  If the platform's size of int is 32 bits, but the size of pointers is 64 bits, the resulting pointer could be truncated as a result of the implicit declaration of malloc() returning a 32-bit integer.

/* #include <stdlib.h> is missing */
 
int main(void) {
  size_t i;
  for (i = 0; i < 100; ++i) {
    char *ptr = (char*)malloc(0x10000000); /* int malloc() assumed */
    *ptr = 'a';
  }
  return 0;
}

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