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Use of plain char type for numeric value

Plain char variable in arithmetic operation without explicit signedness

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Description

This defect occurs when char variables without explicit signedness are used in these ways:

  • To store non-char constants.

  • In an arithmetic operation when the char is:

    • A negative value.

    • The result of a sign changing overflow.

  • As a buffer offset.

char variables without a signed or unsigned qualifier can be signed or unsigned depending on your compiler.

Risk

Operations on a plain char can result in unexpected numerical values. If the char is used as an offset, the char can cause buffer overflow or underflow.

Fix

When initializing a char variable, to avoid implementation-defined confusion, explicitly state whether the char is signed or unsigned.

Extend Checker

A default Bug Finder analysis might not raise this defect when the input values are unknown and only a subset of inputs cause an issue. To check for defects caused by specific system input values, run a stricter Bug Finder analysis. See Extend Bug Finder Checkers to Find Defects from Specific System Input Values.

Examples

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#include <stdio.h>

void badplaincharuse(void)
{
    char c = 200;
    int i = 1000;
    (void)printf("i/c = %d\n", i/c);
}

In this example, the char variable c can be signed or unsigned depending on your compiler. Assuming 8-bit, two's complement character types, the result is either i/c = 5 (unsigned char) or i/c = -17 (signed char). The correct result is unknown without knowing the signedness of char.

Correction — Add signed Qualifier

One possible correction is to add a signed qualifier to char. This clarification makes the operation defined.

#include <stdio.h>

void badplaincharuse(void)
{
    signed char c = -56;
    int i = 1000;
    (void)printf("i/c = %d\n", i/c);
}

Result Information

Group: Numerical
Language: C | C++
Default: Off
Command-Line Syntax: BAD_PLAIN_CHAR_USE
Impact: Medium

Version History

Introduced in R2016b

See Also

Topics