我想知道如何在C99中声明(a const pointer to (a mutable pointer to (a const type)))。
我们说我有这个电话网站：
const uint8_t* result;
create(&result);
void create(const uint8_t * const * resultPtr)是否是在这种情况下声明被调用者的正确方法，还是意味着无意的？我不需要重新分配resultPtr，调用者也不应该[从这个角度来看] *resultPtr[0]，但我的功能需要指定*resultPtr = …。
答案 0 :(得分：4)
我们问cdecl：
$ cdecl declare p as const pointer to pointer to const char
const char ** const p
此声明可以从右到左完全阅读以获得英文版本（记住const char和char const是相同的）。
答案 1 :(得分：1)
一些经验法则：
indexer = 0
for i in df1.columns: # pulls the column name from the above df
for j in df1[i].iteritems(): #pulls the tagid from that row in the df
if i == list(ctrl_data.keys())[indexer] and j[0] == order[0]: # just checks that the loop is iterating correctly
print(j[1])
print(i)
ctrl_data[i]['time_stamp'] = values_pull_process(j[1],start1,end1,num_points,token,base_uri)[0]
ctrl_data[i][order[0]].clear()
print(ctrl_data[i][order[0]])
ctrl_data[i][order[0]] = values_pull_process(j[1],start1,end1,num_points,token,base_uri)[1]
print(ctrl_data[i][order[0]])
elif i == list(ctrl_data.keys())[indexer] and j[0] == order[1]:
ctrl_data[i][order[1]] = values_pull_process(j[1],start1,end1,num_points,token,base_uri)[1]
elif i == list(ctrl_data.keys())[indexer] and j[0] == order[2]:
ctrl_data[i][order[2]] = values_pull_process(j[1],start1,end1,num_points,token,base_uri)[1]
elif i == list(ctrl_data.keys())[indexer] and j[0] == order[3]:
if j[1] == 'NaN':
ctrl_data[i][order[3]] = j[1]
else:
ctrl_data[i][order[3]] = values_pull_ctrl(j[1],start1,end1,num_points,token,base_uri)[1]
elif i == list(ctrl_data.keys())[indexer] and j[0] == order[4]:
if j[1] == 'NaN':
ctrl_data[i][order[4]] = j[1]
else:
ctrl_data[i][order[4]] = values_pull_ctrl(j[1],start1,end1,num_points,token,base_uri)[1]
elif i == list(ctrl_data.keys())[indexer] and j[0] == order[5]:
if j[1] == 'NaN':
ctrl_data[i][order[5]] = j[1]
else:
ctrl_data[i][order[5]] = values_pull_ctrl(j[1],start1,end1,num_points,token,base_uri)[1]
elif i == list(ctrl_data.keys())[indexer] and j[0] == order[6]:
if j[1] == 'NaN':
ctrl_data[i][order[6]] = j[1]
else:
ctrl_data[i][order[6]] = values_pull_process(j[1],start1,end1,num_points,token,base_uri)[1]
elif i == list(ctrl_data.keys())[indexer] and j[0] == order[7]:
ctrl_data[i][order[7]] = j[1]
elif i == list(ctrl_data.keys())[indexer] and j[0] == order[8]:
ctrl_data[i][order[8]] = j[1]
else:
print('Unable to find data for loop: {}, type: {}, tag_id: {}'.format(i,j[0],j[1]))
print('')
indexer = indexer + 1
始终创建指向<type> * const <varName> = <value>内容的不变指针<varName>。请考虑星号<type>右侧的const。* const const <type> <pointers> <varName> = <value> <type> const <pointers> <varName> = <value> const <type> const <pointers> <varName> = <value> 注意：最后两个选项返回
警告：重复的'const'声明说明符[-Wduplicate-decl-specifier]
在Clang编译器上，应该被视为不良做法。
因此，要使指向不变的指针的不可变指针指向<type> const const <pointers> <varName> = <value>，所有这些选项都起作用：
uint8_t const uint8_t * * const result = NULL uint8_t const * * const result = NULL const uint8_t const * * const result = NULL 但是，第二个选项
uint8_t const const * * const result = NULL似乎是恕我直言的最易读和规范的方法。您可能会发现this post在这方面也很有帮助。