函数的返回语句没有按预期工作


return statement of function is not working as expected

我有4个函数在我的php脚本,这将访问升序,在最后一个函数我想返回一个值,但分配不工作。

我的php脚本:
function fourth_f($var){
   $var = $var + 1;
   return $var; //this will make the $var = 6 and return to main process
}
function third_f($var){
   $var = $var + 1;
   fourth_f($var); //this will make the $var = 5
}
function second_f($var){
   $var = $var + 1;
   third_f($var); //this will make the $var = 4
}
function first_f($var_1, $var_2){  
   $var = $var_1 + $var_2 + 1;  
   second_f($var); //this will make the $var = 3
}
//The main
$var_1 = 1;
$var_2 = 1;
$final_var = first_f($var_1, $var_2);
//And i want to echo it here
echo $final_var;

当我执行脚本时,没有错误,但也没有结果,它只是空白,我期待6作为结果。

有人知道什么是错的,我怎么能正确地从最后一个函数返回值?

第四个函数将返回值给第四个函数。因此,您必须返回每个函数中的每个函数调用,以便能够将值分配给变量,例如

function fourth_f($var){
   $var = $var + 1;
   return $var; //returns this value to function third_f
}
function third_f($var){
   $var = $var + 1;
   return fourth_f($var); //returns this value to function second_f
}
function second_f($var){
   $var = $var + 1;
   return third_f($var); //returns this value to function first_f
}
function first_f($var_1, $var_2){  
   $var = $var_1 + $var_2 + 1;  
   return second_f($var); //returns this value to the assignment
}