void value not ignored as it ought to be

今天解决了一个问题:
void value not ignored as it ought to be / return-statement with a value, in function returning 'voi
我想用changeA的结果去影响changeB的if判断.
开始我的方法是建立两个全局变量a和b,试图在changeA函数内部修改全局变量int a的值,结果测试后发现,在changeA里修改后,值可以改变,但是changeA运行结束后int a的值依然为1,并没有发生变化,

int a = 1; int b = 1;void setup() {// put your setup code here, to run once:Serial.begin(115200); }void loop() {// put your main code here, to run repeatedly:changeA(a);changeB(a); }void changeA(int a) {a = 100;Serial.println("在changeA中的a=");Serial.println(a); } void changeB(int a) {if (a == 100){b =100;} }

所以我就在loop中增加了println的监控,我发现a只是在changeA中发生了改变,全局变量一直都没有因为changeA的运行发生变化,所以我改变了我的思路,我准备将changeA实例化给retA, 然后return一个返回值,再将retA传参给changeB进行判断.

int a = 1; int b = 1;void setup() {// put your setup code here, to run once:Serial.begin(115200); }void loop() {// put your main code here, to run repeatedly:Serial.println("在changeA修改前的a=");Serial.println(a);changeA(a);Serial.println("在changeA修改后的a=");Serial.println(a);changeB(a); }void changeA(int a) {a = 100;Serial.println("在changeA中的a=");Serial.println(a); } void changeB(int a) {if (a == 100){b =100;} }

代码如下,修改后编译下发现报错了,void value not ignored as it ought to be 和 return-statement with a value, in function returning ‘void’ [-fpermissive] , 查了资料发现void 建立的函数不能把返回值实例给一个变量,所以就把void改成了int,结果测试成功了.并且我发现changeB参数名称可以和传参的名称不一样

int a = 1; int b = 1;void setup() {// put your setup code here, to run once:Serial.begin(115200); }void loop() {// put your main code here, to run repeatedly:Serial.println("在loop中的a=");Serial.println(a);delay(1000);int retA = changeA(a);Serial.println("在loop中的retA");Serial.println(retA);delay(1000);changeB(retA); }int changeA(int a) {a = 100;return a; } int changeB(int a) {if (a == 100){Serial.println("成功了");} }

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