how to get compile warning

Question

I'm quite surprised when I compile the following code without any warning using g++ 4.1.2 with -Wall -Wextra -Wconversion enabled.

I want g++ to show me every warning to avoid potential harm. I have to stick to g++ 4.1.2.

#include <stdint.h> #include <string> using namespace std; int main() { uint8_t u1=1; uint64_t u64=1000; string s1=""; u1=u64; // want warning here s1=u64; // want warning here s1=u1; }

I think you'll find the English phrase is "want to", not "wanna". — paxdiablo
– paxdiablo, Commented Jul 9, 2010 at 7:42
And you don't want a warning on the last line? What is the intended meaning of assignment from unsigned char to string? — David Rodríguez - dribeas
– David Rodríguez - dribeas, Commented Jul 9, 2010 at 7:52
So you also need -Wsign-conversion. I think it's time for -Wall-but-this-time-for-real. O my. — Philipp
– Philipp, Commented Jul 9, 2010 at 8:06
@Georg: thanks I failed to check for that. I have never used it and did not think that would be supported. — David Rodríguez - dribeas
– David Rodríguez - dribeas, Commented Jul 9, 2010 at 8:28

Philipp · Accepted Answer · 2010-07-09 07:49:01Z

4

I'm afraid GCC before 4.3 doesn't seem to support this. The description of -Wconversion changed between 4.2 and 4.3 to reflect the new warning behavior, and there is no indication that pre-4.3 GCC would check for this.

answered Jul 9, 2010 at 7:49

Philipp

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3 Comments

Anycorn Over a year ago

I can confirm it, gcc 4.3 produces warning, 4.2 does not

rubenvb Over a year ago

I can also confirm GCC 4.4 produces the warning, and only with -Wconversionenabled

lyman Over a year ago

Thx. Then I'll have to look for some static c++ code analysis tool.

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