Gcc does not give any warnings when converting from 64 bits to 32 bits

Question

This is my code:

int main() { uint64_t a = 100; uint32_t b = a; return 0; } 

Gcc command:

 ~$ gcc -o 1 1.c -Wall ~$ 

Can anyone help me?

Answer 1

Use the -Wconversion option.

-Wconversion Warn about implicit conversions that may change the value. This includes conversions between real and integer, such as abs (x) when x is double; Conversions between signed and unsigned, for example unsigned ui = -1; and conversions to smaller types, for example sqrtf (M_PI). Do not warn about explicit casts such as abs ((int) x) and ui = (unsigned) -1, or if the value does not change during conversion, as in abs (2.0). Conversion warnings between unsigned and unsigned integers can be disabled using -Wno-sign-conversion.

For C ++, also warn of confusing overload resolution for custom conversions; and conversions that never use the type conversion operator: conversions to void, the same type, base class, or a reference to them. Conversion warnings between signed and unsigned integers are disabled by default in C ++ unless -Wsign-conversion is explicitly enabled.

In your code:

converting to "uint32_t {aka unsigned int}" from "uint64_t {aka long long unsigned int}" may change its value [-Wconversion]

Answer 2

It is a common misconception that -Wall includes all warnings.

It includes “all design warnings that some users find dubious, which are easy to avoid (or modify to prevent warnings) even when combined with macros” (citing the GCC manual).

Even -Wextra only "includes some additional warning flags that are not activated by -Wall" (again from the GCC manual).

There is also -pedantic , which generates warnings in cases where the meaning of the code is clear to the compiler, but the standard requires that the corresponding compiler produce a message. (By default, GCC needs to continue compiling).

Even if all three are turned on, you will not receive every warning that the compiler can provide. See the compiler manual for more details.