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The C C23 Standard requires type specifiers and forbids implicit function declarations. The C90 Standard allows for implicit typing of variables and functions. Because implicit declarations lead to less stringent type checking, they can often introduce unexpected and erroneous behavior or even security vulnerabilities.The C Standard requires type identifiers and forbids implicit function declarations. After issuing the diagnostic, Consequently, some existing legacy code uses implicit typing. Some C compilers still support legacy code by allowing implicit typing, but it should not be used for new code. Such an implementation may choose to assume an implicit declaration and continue translation to support existing programs that used this feature.

Noncompliant Code Example (Implicit int)

C no longer allows for the absence of type specifiers in a declaration. In these cases, the type is defined to be that of a signed int.Do not rely on implicit int typing. Section 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;

Most Some C implementations do not issue a diagnostic for the violation of this constraint. Many These nonconforming C translators will 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.:

extern int foo;

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Implicit declaration of functions is not allowed: ; every function must be explicitly declared before it can be called. In C89C90, if a function is called without an explicit prototype, the compiler provides an implicit declaration.

The C90 Standard Standard [ISO/IEC 9899:1990] includes this requirement:

If the expression that precedes the parenthesized argument list in a function call consists solely of an identifier, and if no declaration is visible for this identifier, the identifier is implicitly declared exactly as if, in the innermost block containing the function call, the declaration extern int identifier(); appeared.

A C99 and later implementation will not perform implicit function declarations.

If a function declaration is not visible at the point at which a call to the function is made, some compilers 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 return an int. However, to conform with to the current C Standard, you programmers must explicitly prototype every function before invoking it. This noncompliant example fails to prototype the foo() function before invoking it in main() An implementation that conforms to the C Standard may or may not perform implicit function declarations, but C does require a conforming implementation to issue a diagnostic if it encounters an undeclared function being used.

In 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(). If the platform's size of int is 32 bits, but the size of pointers is 64 bits, the resulting pointer would likely be truncated as a result of the implicit declaration of malloc(), returning a 32-bit integer.

#include <stddef.h>
/* #include <stdlib.h> is missing */
 
int main(void) {
  for int c = foo();
  printf("%d\n", c(size_t i = 0; i < 100; ++i) {
    /* int malloc() assumed */
    char *ptr = (char *)malloc(0x10000000);
  return 0;
}

int foo(int a) {  *ptr = 'a';
  }
  return a0;
}

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Compliant Solution (Implicit Function Declaration)

In this This compliant solution , a prototype for foodeclares malloc() appears before the function invocation. by including the appropriate header file:

int foo(int);
#include <stdlib.h>
 
int main(void) {
  int cfor (size_t i = foo(0);
  printf("%d\n", c);
  return 0;
}

int foo(int a) {0; i < 100; ++i) {
    char *ptr = (char *)malloc(0x10000000);
    *ptr = 'a';
  }
  return a0;
}

For more information on function declarations, see 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 explicily 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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ISO/IEC TR 24772 "OTR Subprogram signature mismatch"

MISRA

Sources

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