Linux应用程序控制GPIO，给个简单的完整操作范例

dzjsxylmj 2011-06-17 03:37:47

s3c2440 FL开发板，想不写GPIO驱动程序，直接在应用程序中驱动GPIO，求一个简单可行的操作范例，谢谢。我得内核版本是2.6.31.

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booboo18 2012-05-05

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楼主的问题不知道解决没有呢？我也想用个驱动GPIO口的应用程序，驱动要是内核带了得话，不是应该写个驱动就行了么？

丰满 2011-07-20

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恩，是可以在应用层做的，有知道gpio的真实的物理地址，映射，呵呵，那还不如自己写个gpio的驱动的了，多简单啊

tian0xin0 2011-07-13

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直接在应用层写驱动，你需要知道控制GPIO灯的寄存器地址。一般的话在开发板的使用手册里面都应该有。下面的代码是研祥工控主板上应用层直接控制GPIO口的代码：
#define INDEX_PORT 0x2E
#define DATA_PORT 0x2F

void GPIOControl::GPIOInitialize()
{
//comment out to test code on platform without GPIO
unsigned char tmp_val;
if(ioperm(INDEX_PORT,2,1))
{
printf（"ioperm error!\n"）;
}
outb(0x87, INDEX_PORT);
outb(0x87,INDEX_PORT);
outb(0x29,INDEX_PORT);
tmp_val = (inb(DATA_PORT )|0x02)&0xfb;
outb(tmp_val, DATA_PORT);
outb(0x2c, INDEX_PORT);
tmp_val = inb(DATA_PORT ) &0x7f;
outb(tmp_val, DATA_PORT);
outb(0x07, INDEX_PORT);
outb( 0x09, DATA_PORT);
outb(0x30 ,INDEX_PORT);
outb(0x03,DATA_PORT);
outb(0xe3, INDEX_PORT);
tmp_val = inb(DATA_PORT) & 0xfc;
outb(tmp_val, DATA_PORT);
outb( 0xf0 ,INDEX_PORT);
tmp_val = (inb(DATA_PORT)& 0x3f) | 0x33;
outb(tmp_val, DATA_PORT);
}

void GPIOControl::Out_Lev(int pin, int lev_val)
{
//comment out to test code on platform without GPIO
GPIOInitialize();
unsigned char oldval;
int pin_val[4] = {36, 37, 20, 21};
unsigned int reg_val[4] = {0xf1, 0xf1, 0xe4, 0xe4};
outb(0x07,INDEX_PORT);
outb(0x09, DATA_PORT);
outb(reg_val[pin - 1] +1,INDEX_PORT);
oldval = inb(DATA_PORT) & ( ~ (0x01 << pin_val[pin -1] % 10));
outb(oldval,DATA_PORT);
outb(reg_val[pin -1],INDEX_PORT);
oldval = inb(DATA_PORT);
oldval = lev_val? oldval|(0x01 << pin_val[pin -1]%10): oldval& (~ (0x01 << pin_val[pin-1] % 10));
outb( oldval,DATA_PORT);
}