试图理解函数指针的用法 [英] Trying to understand the usage of function pointer

问题描述

这是在 U中的作用从那里内核启动-boot的bootm.c ：

/* Subcommand: GO */
static void boot_jump_linux(bootm_headers_t *images, int flag)
{
#ifdef CONFIG_ARM64
void (*kernel_entry)(void *fdt_addr);
int fake = (flag & BOOTM_STATE_OS_FAKE_GO);

kernel_entry = (void (*)(void *fdt_addr))images->ep;

debug("## Transferring control to Linux (at address %lx)...\n",
(ulong) kernel_entry);
bootstage_mark(BOOTSTAGE_ID_RUN_OS);

announce_and_cleanup(fake);

if (!fake)
kernel_entry(images->ft_addr);
#else
unsigned long machid = gd->bd->bi_arch_number;
char *s;
void (*kernel_entry)(int zero, int arch, uint params);
unsigned long r2;
int fake = (flag & BOOTM_STATE_OS_FAKE_GO);

kernel_entry = (void (*)(int, int, uint))images->ep;

s = getenv("machid");
if (s) {
strict_strtoul(s, 16, &machid);
printf("Using machid 0x%lx from environment\n", machid);
}

debug("## Transferring control to Linux (at address %08lx)" \
"...\n", (ulong) kernel_entry);
bootstage_mark(BOOTSTAGE_ID_RUN_OS);
announce_and_cleanup(fake);

if (IMAGE_ENABLE_OF_LIBFDT && images->ft_len)
r2 = (unsigned long)images->ft_addr;
else
r2 = gd->bd->bi_boot_params;

if (!fake)
kernel_entry(0, machid, r2);
#endif
}

我对着了解如何 kernel_entry 正在这里工作的难度。特别是在第二最后一行它被用作

I am facing difficulty in understanding how kernel_entry is working here. Especially in the second-last line it is being used as:

   kernel_entry(0, machid, r2);

那么，是 kernel_entry的定义（）？我失败了整个U-boot和内核源代码code找到。

So where is the definition of kernel_entry()? I failed to find in entire u-boot and kernel source code.

更新

我在这里改写我的问题：

I am rephrasing my question here:

假设 kernel_entry 是指向一个功能被定义为：

Suppose kernel_entry is a pointer to a function and is being defined as:

bootm_headers_t *images
kernel_entry = (void (*)(int, int, uint))images->ep;

然后程序中某个位置它被称为：

Then somewhere in the program it is being called as:

 kernel_entry(0, machid, r2);

我理解所的指针， kernel_entry 应存储函数的地址。但我想了解什么操作都将针对三个参数进行。为什么我们这些争论？

I understand being a pointer, kernel_entry should store an address of a function. But I want to understand what operations will be performed on the three arguments. Why do we have those arguments?

推荐答案

kernel_entry 的声明变量及其类型，它是采用指针给一个函数 INT ， INT ， UINT 并返回无效（可能是最令人困惑的部分），是在这里：

The declaration of kernel_entry variable and its type, which is a pointer to a function taking int, int, uint and returning void (probably the most confusing part), is here:

void (*kernel_entry)(int zero, int arch, uint params);

任务，图像 - ＆GT; EP 被强制转换成所需的签名函数指针并投入变量：

Assignment, images->ep is cast into desired signature function pointer and put into the variable:

kernel_entry = (void (*)(int, int, uint))images->ep;

最后，调用该函数：

Finally, the function is called:

kernel_entry(0, machid, r2);

请注意，如果CONFIG_ARM64定义，则函数 kernel_entry 点有不同的签名：

Please note that if CONFIG_ARM64 is defined, then the function kernel_entry points to has different signature:

void (*kernel_entry)(void *fdt_addr); //takes one void* param and returns void

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