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The C Standard does not allow for implicit typing of variables and functionsC23 Standard requires type specifiers and forbids implicit function declarations. The C90 Standard did allow such allows implicit typing of variables and functions. Consequently, there exists some existing legacy code that uses implicit typing. Some C compilers still support legacy code by allowing implicit typing, but it should not be used for new code. Because implicit declarations lead to less stringent type checking, they can introduce unexpected and erroneous behavior or even security vulnerabilities.The C Standard requires type identifiers and forbids implicit function declarations. After issuing the diagnostic, Such an implementation may choose to assume an implicit declaration and continue translation to support existing programs that used this feature.

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C no longer allows the absence of type specifiers in a declaration.  Subclause The C Standard, 6.7.3 paragraph 2 of the C Standard  [ ISO/IEC 9899:20112024 ], states:

Except where the type is inferred (6.7.10), at At least one type specifier shall be given in the declaration specifiers in each declaration, and in the specifier-qualifier list in each struct member declaration and type name.

This noncompliant code example omits the type specifier:

extern foo;

Some C implementations do not issue a diagnostic for the violation of this constraint. These noncompliant nonconforming C translators continue to treat such declarations as implying the type int.

Compliant

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Solution (Implicit int)

This compliant solution explicitly includes a type specifier:

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Implicit declaration of functions is not allowed: ; every function should must be explicitly declared before it can be called. In C90, if a function is called without an explicit prototype, the compiler provides an implicit declaration.

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If a function declaration is not visible at the point at which a call to the function is made, C90-compliant platforms assume an implicit declaration of extern int funcidentifier();.

This declaration implies that the function may take any number and type of arguments and returns a single return an int. However, to conform with to the current C Standard, you programmers must explicitly prototype every function before invoking it. An implementation that conforms to the C Standard may or may not perform implicit function declarations. However, but C does require the a conforming implementation to issue a diagnostic if it encounters an undeclared function being used.

In the following this noncompliant code example, if malloc() is not declared, either explicitly or by including stdlib.h, a compiler that conforms only to C90 may implicitly declare malloc() as int malloc(). (Compilers that comply only 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 would likely be truncated as a result of the implicit declaration of malloc(), returning a 32-bit integer.

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#include <stddef.h>
/* #include <stdlib.h> is missing */
 
int main(void) {
  for (size_t i

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= 0; i < 100; ++i) {
    /* int malloc() assumed */
    char *ptr = (char *)malloc(0x10000000);

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    *ptr = 'a';
  }
  return 0;
}

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This compliant solution declares malloc() by including the appropriate header file.:

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

For more information on function declarations, see DCL07-C. Include the appropriate type information in function declarators.

Noncompliant Code Example (Implicit Return Type)

Similarly, do Do not declare a function with implicit an implicit return type. If it returns For example, if a function returns a meaningful integer value, declare it it as returning int. If it returns no meaningful value, declare it it as returning void.

#include <limits.h>
#include <stdio.h>
 
foo(void) {
  return UINT_MAX;
}

int main(void) {
  long long int c = foo();
  printf("%lld\n", c);
  return 0;
}

Because the compiler assumes that foo() returns a value of type int for this noncompliant code example, UINT_MAX is incorrectly converted to −1.

Compliant Solution (Implicit Return Type)

This compliant solution explicitly defines the return type of foo() as unsigned int:. As a result, the function correctly returns UINT_MAX .

#include <limits.h>
#include <stdio.h>

unsigned int foo(void) {
  return UINT_MAX;
}

int main(void) {
  long long int c = foo();
  printf("%lld\n", c);
  return 0;
}

Risk Assessment

Because implicit declarations lead to less stringent type checking, they can introduce unexpected and erroneous behavior. Occurrences of an omitted type specifier in existing code are rare, and the consequences are generally minor, perhaps resulting in abnormal program termination.

Automated Detection

Related Vulnerabilities

Search for vulnerabilities resulting from the violation of this rule on the CERT website.

Related Guidelines

Key here (explains table format and definitions)

Bibliography

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