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");
}
}
