Question

有些人在使用make命令进行编译时遇到了问题，所以我想我会尝试使用以下操作系统的ubuntu 32位并挤压64位

我克隆了git项目https://github.com/duckduckgo/cpp-libface/运行了make命令，我不断收到以下内容

    g++ -o lib-face src/main.cpp src/httpserver.o deps/libuv/libuv.a deps/http-parser/http_parser.o -I . -I deps -Wall  -D_FILE_OFFSET_BITS=64 -O2 -lm -lrt -pthread
In file included from src/main.cpp:21:
./include/benderrmq.hpp: In member function 'uint_t LookupTables::query_max(uint_t, uint_t, uint_t)':
./include/benderrmq.hpp:250: warning: format '%d' expects type 'int', but argument 4 has type 'size_t'
./include/benderrmq.hpp:251: warning: format '%d' expects type 'int', but argument 4 has type 'size_t'
./include/benderrmq.hpp:252: warning: format '%d' expects type 'int', but argument 4 has type 'size_t'
./include/benderrmq.hpp: In member function 'void BenderRMQ::initialize(const vui_t&)':
./include/benderrmq.hpp:353: warning: format '%d' expects type 'int', but argument 3 has type 'size_t'
./include/benderrmq.hpp:353: warning: format '%d' expects type 'int', but argument 4 has type 'size_t'
./include/benderrmq.hpp:354: warning: format '%d' expects type 'int', but argument 3 has type 'size_t'
./include/benderrmq.hpp:354: warning: format '%d' expects type 'int', but argument 4 has type 'size_t'

日志更长，但只是拉出重要部分

错误的文件位于https://raw.githubusercontent.com/duckduckgo/cpp-libface/master/include/benderrmq.hpp

Answer 1

这是printf的标准警告。好的开始是：

man 3 printf

基本上解决方案就是：

std::size_t a0 = 1234;
printf("%zu\n", a0);

而不是：

std::size_t a0 = 1234;
printf("%d\n", a0);

Answer 2

您在此处定义的断言宏：

#define assert_lt(X,Y) if (!((X)<(Y))) { fprintf(stderr, "%d < %d FAILED\n", (X), (Y)); assert((X)<(Y)); }
#define assert_gt(X,Y) if (!((X)>(Y))) { fprintf(stderr, "%d > %d FAILED\n", (X), (Y)); assert((X)>(Y)); }
#define assert_le(X,Y) if (!((X)<=(Y))) { fprintf(stderr, "%d <= %d FAILED\n", (X), (Y)); assert((X)<=(Y)); }
#define assert_eq(X,Y) if (!((X)==(Y))) { fprintf(stderr, "%d == %d FAILED\n", (X), (Y)); assert((X)==(Y)); }
#define assert_ne(X,Y) if (!((X)!=(Y))) { fprintf(stderr, "%d != %d FAILED\n", (X), (Y)); assert((X)!=(Y)); }

https://github.com/duckduckgo/cpp-libface/blob/master/include/utils.hpp

使用"%d"中用于整数的fprintf格式字符串。

但是你将宏用于在这里返回size_t的方法：

assert_lt(index, repr.size());
assert_lt(l, repr[0].size() + 1);
assert_lt(u, repr[0][0].size());

在这里：

assert_eq(levels.size(), euler.size());
assert_eq(levels.size(), rev_mapping.size());

https://github.com/duckduckgo/cpp-libface/blob/master/include/benderrmq.hpp

因此，您可以创建使用"%zu"的其他宏，也可以将size方法的结果转换为有符号整数（如果您知道很可能没有溢出问题）。

使用benderrmq犯错误

2 个答案: