[英]error “cannot appear in a constant-expression” in g++ but not in gcc
98 static inline int set_hw_br(pid_t tracee, dr7_t *pdr7, void *addr, int dr_index)
99 {
100 errno = 0;
101 printf("LINE = %d <pid> %d, dr_index= %d addr=%u \n",__LINE__, tracee, dr_index, addr);
102 //if (ptrace(PTRACE_POKEUSER, tracee, offsetof(struct user, u_debugreg[dr_index]), addr))
103 if (ptrace(PTRACE_POKEUSER, tracee, offsetof(struct user, u_debugreg[dr_index]), addr))
104 {
105 int ii = errno;
106 printf("MKH: 22 errno = %d\n", ii);
107 ptrace(PTRACE_DETACH, tracee, 0, 0);
108 return -1;
109 }
110 else
111 printf("PTRACE_POKEUSER passed...\n");
Above code(part of main code) is successfully compiled in GCC compiler. 以上代码(部分主要代码)已在GCC编译器中成功编译。 But while compiling through G++, it is giving fillowing error:
error: 'dr_index' cannot appear in a constant-expression
in line 103. set_hw_br is called from another function. 但是,当通过G ++进行编译时,它却出现了错误:
error: 'dr_index' cannot appear in a constant-expression
在第103行error: 'dr_index' cannot appear in a constant-expression
中。set_hw_br是从另一个函数调用的。
Any idea why this failing in g++? 知道为什么这在g ++中会失败吗?
Thanks. 谢谢。
The offsetof
macro requires that the member-designator has to produce an address constant (C11 7.19/3): offsetof
宏要求成员标识符必须产生一个地址常数(C11 7.19 / 3):
offsetof(type, member-designator)
which expands to an integer constant expression that has type
size_t
, the value of which is the offset in bytes, to the structure member (designated by member-designator ), from the beginning of its structure (designated by type ).它扩展为一个整数常量表达式,其类型为
size_t
,其值是从结构的开头(由type 指定 )到结构成员(由member-designator指定 )的偏移量(以字节为单位)。 The type and member designator shall be such that given类型和成员代号应使
static type t;
then the expression
&(t.
member-designator)
evaluates to an address constant.然后表达式
&(t.
-designator)
得出一个地址常数。 (If the specified member is a bit-field, the behavior is undefined.)(如果指定的成员是位字段,则行为未定义。)
In your code, t.u_subreg[dr_index]
is not a constant, because dr_index
is not a constant. 在您的代码中,
t.u_subreg[dr_index]
不是常数,因为dr_index
不是常数。
GCC implements offsetof
with a compiler intrinsic so what is allowed in an offsetof
expression depends on the rules of GCC's intrinsic. GCC使用编译器内部函数实现
offsetof
,因此offsetof
表达式中允许的内容取决于GCC内部函数的规则。 As an extension to the standard, the GCC C front-end allows a non-constant expression as input and produces a non-constant result. 作为标准的扩展,GCC C前端允许使用非恒定表达式作为输入并产生非恒定结果。 The C++ front-end does not allow it, giving the error telling you
dr_index
cannot be used there. C ++前端不允许这样做,并给出错误消息,告诉您
dr_index
无法在此处使用。
You can change the offsetof
expression to only use constants: 您可以将
offsetof
表达式更改为仅使用常量:
offsetof(struct user, u_debugreg[0])
then you can add the index to it, where T
is the type in the array u_debugreg
: 然后可以向其添加索引,其中
T
是数组u_debugreg
的类型:
offsetof(struct user, u_debugreg[0]) + sizeof(T)*dr_index
(This assumes that u_debugreg
is an actual array, not a pointer). (这假设
u_debugreg
是实际数组,而不是指针)。
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