我有以下代码，但我觉得很脏..
我不想写这么多ifs，然后在每个if中重复输入代码..
关于如何改进此代码的任何想法？
char obj[5];
strlcpy(obj, &jarr[i], arr[i]);
if( !strcmp( obj, "led_r" ) ){
i++;
strlcpy( obj, &jarr[i], arr[i] );
red_brightness = atoi( obj );
Serial.print(" RED: ");
Serial.println( red_brightness );
}
if( !strcmp( obj, "led_g" ) ){
i++;
strlcpy( obj, &jarr[i], arr[i] );
green_brightness = atoi( obj );
Serial.print(" GREEN: ");
Serial.println( green_brightness );
}
if( !strcmp( obj, "led_b" ) ){
i++;
strlcpy( obj, &jarr[i], arr[i] );
blue_brightness = atoi( obj );
Serial.print(" BLUE: ");
Serial.println( blue_brightness );
}
答案 0 :(得分：5)
在此之前的其他答案也很好。这个答案的优点是你可以将选项名称更改为任何东西，它仍然可以工作。您还可以将此类选项解析扩展为更大的选项集。
// Assuming these are globals.
int red_brightness, green_brightness, blue_brightness;
// Use this array to help you parse.
static const struct {
const char *optionName;
int *brightness;
const char *label;
} ledOptions[] = {
{ "led_r", &red_brightness, " RED: " },
{ "led_g", &green_brightness, " GREEN: " },
{ "led_b", &blue_brightness, " BLUE: " },
};
// A handy macro for later.
#define DIM(array) (sizeof(array) / sizeof(array[0]))
...
// Now in your actual code:
strlcpy(obj, &jarr[i], arr[i]);
for (j=0;j<DIM(ledOptions);j++) {
if( !strcmp( obj, ledOptions[i].optionName ) ){
i++;
strlcpy( obj, &jarr[i], arr[i] );
*ledOptions[i].brightness = atoi( obj );
Serial.print(ledOptions[i].label);
Serial.println(*ledOptions[i].brightness);
}
}
答案 1 :(得分：2)
将前4个字符与“led_”进行比较，确保其长度为5，然后打开第五个字符。将其余代码重构为函数。
答案 2 :(得分：2)
公共代码可以从if语句
中计算出来您可以将switch语句用于字符串
中唯一更改的字符switch(obj[4]) {
case 'r':
Serial.println("RED: " + red_brightness );
case 'g':
Serial.println("GREEN: " + green_brightness );
case 'b':
Serial.println("BLUE: " + blue_brightness );
}
答案 3 :(得分：0)
这就是我想出的。免责声明，我还没有尝试编译/运行它。
for (int i=0; i<6; i++)
{
i++;
int b = atoi( &jarr[i] );
switch (i)
{
case 1:
Serial.print(" RED: ");
red_brightness = b;
break;
case 3:
Serial.print(" GREEN: ");
green_brightness = b;
break;
case 5:
Serial.print(" BLUE: ");
blue_brightness = b;
break;
default:
break;
}
Serial.println(b);
}
想要打印的值是在奇数索引处的＆amp; jarr []中。
答案 4 :(得分：0)
这里只是为了好玩我的个人方法...
#ifndef ARDUINO
#include <string.h>
#include <stdio.h>
/* PART of code derived from question but unspecified in code . done to be able to compile */
// this is not specified in question
// either it a char jarr[1024] but then why do we iterate on char ( i++ would go to next char and not to next word )
// either it is a char* jar[6] but then we access its char * content with jarr[i] and not with &jarr[i]
//char * jarr[6];
// this look like to be size of each argument.
//int arr[6];
/*
TEST vectors
*/
char * jarr0[] = { "led_r", "200", "led_g", "100", "led_b", "150" };
int arr0[] = { 5, 3, 5,3, 5,3 };
char * jarr1[] = { "led_g", "200", "led_r", "100", "led_b", "150" };
int arr1[] = { 5, 3, 5,3, 5,3 };
char * jarr2[] = { "lgd_g", "200", "ged_r", "100", "led_b", "150" };
int arr2[] = { 5, 3, 5,3, 5,3 };
int red_brightness = 0;
int blue_brightness = 0;
int green_brightness = 0;
/* end of unspecified PART */
#endif
/*real code *
/* use indirection */
struct led_color_map {
int * color_brightness_p;
const char * color_name;
char key;
} led_color[] = {
{&red_brightness,"RED",'r'},
{&green_brightness,"GREEN",'g'},
{&blue_brightness,"BLUE",'b'}
};
enum { RED_IDX=0, GREEN_IDX, BLUE_IDX };
// for perf test
int miss = 0;
#ifdef ARDUINO
void led_setup()
#else
void led_setup(char * jarr[], int arr[])
#endif
{
char* obj=NULL;
int i=0;
int offset=0;
for (i=0; i <6; i+=2)
{
obj=jarr[i];
// test 'led_' prefix with reverse statistical letter apparence
// which is 'l' then 'd' then 'e' , guessing that '_' can be used in other words ...
if ( ( obj[0] == 'l' ) && ( obj[2] == 'd' ) && (obj[1] == 'e' ) && (obj[3] == '_') )
{
int idx=0;
for ( idx = 0; idx <= BLUE_IDX; idx ++)
{
int rolling_idx = (idx + offset) % ( BLUE_IDX + 1 );
struct led_color_map * color_found = &led_color[rolling_idx];
if ( color_found->key == obj[4] )
{
char ivalue[32];
if ( arr[i+1] < sizeof(ivalue) )
{
#ifdef ARDUINO
strlcpy(ivalue, jarr[i+1], arr[i+1]);
#else
strncpy(ivalue, jarr[i+1], arr[i+1]);
#endif
{
int brightness = atoi( ivalue );
*(color_found->color_brightness_p) = brightness;
#ifdef ARDUINO
Serial.print( color_found->color_name );
Serial.println( *color_brightness );
#else
printf(" %s :%i\n", color_found->color_name, brightness);
#endif
}
offset=(rolling_idx + 1) % ( BLUE_IDX + 1 ); // try with next color first next time.
break;
}
}
else
{
miss ++;
printf("miss %i wrong color order expected %s\n",miss, led_color[rolling_idx].color_name);
}
}
}
else
{
miss ++;
printf("miss %i wrong variable expected led_%c\n", miss, led_color[offset % ( BLUE_IDX + 1 )].key);
}
}
}
#ifndef ARDUINO
int main(int argc, char ** argv)
{
led_setup(jarr0, arr0);
printf("test0 : miss %i\n",miss);
miss=0;
led_setup(jarr1, arr1);
printf("test1 : miss %i\n",miss);
miss=0;
led_setup(jarr2, arr2);
printf("test2 : miss %i\n",miss);
}
#endif
gcc led.c ./a.out
RED :200
GREEN :100
BLUE :150
test0 : miss 0
miss 1 wrong color order expected RED
GREEN :200
miss 2 wrong color order expected BLUE
RED :100
miss 3 wrong color order expected GREEN
BLUE :150
test1 : miss 3
miss 1 wrong variable expected led_r
miss 2 wrong variable expected led_r
miss 3 wrong color order expected RED
miss 4 wrong color order expected GREEN
BLUE :150
test2 : miss 4
答案 5 :(得分：0)
char obj[5];
strlcpy(obj, &jarr[i], arr[i]);
int *brightness;
String ostr();
if( !strcmp( obj, "led_r" ) ){
ostr = " RED: ";
brightness = red_brightness;
}
if( !strcmp( obj, "led_g" ) ){
ostr = " GREEN: ";
brightness = green_brightness;
}
if( !strcmp( obj, "led_b" ) ){
ostr = " BLUE: ";
brightness = blue_brightness;
}
i++;
strlcpy( obj, &jarr[i], arr[i] );
*brightness = atoi( obj );
Serial.print(ostr);
Serial.println(*brightness);