问题描述

怎么样让不同变量重复一个动作？

举例说明我想实现的，比如我有十个变量（整数值）以及变量的值，一个字符串。

然后代码中是判断云的数量决定天气状况：

 if (hour1cloud <= 5) {            hour1weather = @""Clear"";        }        if (5 < hour1cloud <= 25) {            hour1weather = @""Mostly Clear"";        }        if (25 < hour1cloud <= 50) {            hour1weather = @""Partly Cloudy"";        }        if (50 < hour1cloud <= 83) {            hour1weather = @""Mostly Cloudy"";        }        if (83 < hour1cloud <= 105) {            hour1weather = @""Overcast"";        }

然后 hour2cloud hour3cloud hour4cloud分别对应hour2weather hour3weather等。当我输入hour1cloud获取hour1weather的值，能不能对所有组都变成通用方法？

解决方案

方法如下：

- (NSString*)weatherStringFromCloud:(int)cloud {    NSString *weather;    if (cloud <= 5) {        weather = @""Clear"";    } else if (cloud <= 25) {        weather = @""Mostly Clear"";    } else if (cloud <= 50) {        weather = @""Partly Cloudy"";    } else if (cloud <= 83) {        weather = @""Mostly Cloudy"";    } else if (cloud <= 105) {        weather = @""Overcast"";    } else {        weather = nil;    }    return weather;}

然后用不同变量调用：

hour1weather = [self weatherStringFromCloud:hour1cloud];hour2weather = [self weatherStringFromCloud:hour2cloud];hour3weather = [self weatherStringFromCloud:hour3cloud];hour4weather = [self weatherStringFromCloud:hour4cloud];

解决方案二：

static NSString *stringForCloudiness(int cloudiness) {    static int const kCloudinesses[] = { 5 25 50 83 105 };    static NSString *const kStrings[] = { @""Clear"" @""Mostly Clear"" @""Partly Cloudy"" @""Mostly Cloudy"" @""Overcast"" };    static int const kCount = sizeof kCloudinesses / sizeof *kCloudinesses;    for (int i = 0; i < kCount; ++i) {        if (cloudiness <= kCloudinesses[i]) {            return kStrings[i];        }    }    return @""A cloudiness level unparalleled in the history of recorded weather"";}

接下来要同步数组：

static NSString *stringForCloudiness(int cloudiness) {    typedef struct {        int cloudiness;        __unsafe_unretained NSString *string;    } CloudStringAssociation;    static CloudStringAssociation const kAssociations[] = {        { 5 @""Clear"" }        { 25 @""Mostly Clear"" }        { 50 @""Partly Cloudy"" }        { 83 @""Mostly Cloudy"" }        { 105 @""Overcast"" }        { INT_MAX @""A cloudiness level unparalleled in the history of recorded weather"" }    };    int i = 0;    while (cloudiness > kAssociations[i].cloudiness) {        ++i;    }    return kAssociations[i].string;}