Synopsis

An expression is implicitly converted to a narrower type, resulting in an overflow or underflow of its value.

Enabled by default

Yes

Severity/Certainty

Medium/High

Full description

An expression is implicitly converted to a narrower type, resulting in an overflow or underflow of its value. This might be unintended and can cause logic errors. Because unexpected behavior is much more likely than an application crash, such errors can be very hard to find.

Coding standards

Code examples

The following code example fails the check and will give a warning:

typedef int I;
typedef I J;

void f(){
  J x = 375;
  char c = x;  //overflows to 120
}

The following code example passes the check and will not give a warning about this issue:

void f(){
  int x = 35;
  char c = x;
}