类型转换 : lvalue required as left operand of assignment

Question

我只是在没有完整内存管理的系统上使用此代码:

typedef unsigned short component_t;
typedef struct {
    component_t* c;    // least-significant word first
    unsigned int num_components; // number of unsigned short rows
} integer;

integer result;
result.c=malloc(component_t*)malloc(sizeof(component_t)*128); //this is just an example to tell I'm correctly initializing. There's no malloc nor memset inside OpenCL. currently, I'm trying to speed up the code serially before switching to OpencL. That's also why I don't use GMP.
result.num_components=128

下一行:

for(int i=0;i<result.num_components/4;i++) (unsigned long)result.c[i] = 26; // assign 26 in that part of the memory

触发:

gcc integer.c
integer.c:567:36 error: lvalue required as left operand of assignment
     (unsigned long)result.c[5] = 26;
                                ^

我不知道该行的真正问题以及我到底需要写什么来纠正这个问题。
注意:我也见过(即使这没用):

for(int i=0;i<result.num_components/4;i++) (unsigned long)result.c[i]++

编译时

for(int i=0;i<result.num_components/4;i++) (unsigned long)result.c[i]+=1

没有。

Answer 1

integer.c:567:36 error: lvalue required as left operand of assignment
     (unsigned long)result.c[5] = 26;
                                ^

result.c[5]的值被检索并转换为unsigned long ，这会给您留下类似于 15 = 26 的分配。您(大概)想要的是转换 result.c指向 unsigned long 的指针 :

 ((unsigned long *)result.c)[5] = 26;