I would simplify this by introducing a variable max:
public static int max(int a, int b, int c, int d) {
int max = a;
if (b > max)
max = b;
if (c > max)
max = c;
if (d > max)
max = d;
return max;
}
You could also use Math.max, as suggested by fast snail, but since this seems to be homework, I would prefer the algorithmic solution.
Math.max(Math.max(a,b),Math.max(c,d))
Answer from Patrick Hofman on Stack OverflowVideos
I would simplify this by introducing a variable max:
public static int max(int a, int b, int c, int d) {
int max = a;
if (b > max)
max = b;
if (c > max)
max = c;
if (d > max)
max = d;
return max;
}
You could also use Math.max, as suggested by fast snail, but since this seems to be homework, I would prefer the algorithmic solution.
Math.max(Math.max(a,b),Math.max(c,d))
Try Math.max like below:
return Math.max(Math.max(a, b), Math.max(c, d));
int
public static int min(int a, int b) {
return (a <= b) ? a : b;
}
long
public static long min(long a, long b) {
return (a <= b) ? a : b;
}
float
public static float min(float a, float b) {
if (a != a) return a; // a is NaN
if ((a == 0.0f) && (b == 0.0f) && (Float.floatToIntBits(b) == negativeZeroFloatBits)) {
return b;
}
return (a <= b) ? a : b;
}
double
public static double min(double a, double b) {
if (a != a) return a; // a is NaN
if ((a == 0.0d) && (b == 0.0d) && (Double.doubleToLongBits(b) == negativeZeroDoubleBits)) {
return b;
}
return (a <= b) ? a : b;
}
More info: Here
Java 7 documentation:
Returns the smaller of two int values. That is, the result the argument closer to the value of Integer.MIN_VALUE. If the arguments have the same value, the result is that same value.
Behaviour:
Math.min(1, 2) => 1
Math.min(1F, 2) => 1F
Math.min(3D, 2F) => 2D
Math.min(-0F, 0F) => -0F
Math.min(0D, -0D) => -0D
Math.min(Float.NaN, -2) => Float.NaN
Math.min(-2F, Double.NaN) => Double.NaN
java.lang.Math and java.lang.StrictMath Source:
public static int min(int a, int b) {
return (a <= b) ? a : b;
}
java.lang.Math Bytecode (javap -c Math.class of Oracle's JDK's JRE's rt.jar):
public static int min(int, int);
Code:
0: iload_0 // loads a onto the stack
1: iload_1 // loads b onto the stack
2: if_icmpgt 9 // pops two ints (a, b) from the stack
// and compares them
// if b>a, the jvm continues at label 9
// else, at the next instruction, 5
// icmpgt is for integer-compare-greater-than
5: iload_0 // loads a onto the stack
6: goto 10 // jumps to label 10
9: iload_1 // loads
10: ireturn // returns the currently loaded integer
If the comparison at 5 is true, a will be loaded, the jvm will jump to 10 and return a, if the comparison yields false, it will jump to 9, which will load and return b.
Intrinsics:
This .hpp file of the Java 8 Hotspot JVM hints that it optimizes Math.min even further with optimized machine code:
do_intrinsic(_min, java_lang_Math, min_name, int2_int_signature, F_S)
This means the above bytecode won't be executed by the Java 8 Hotspot JVM. However, this differs from JVM to JVM, which is why I also explained the bytecode!
Hopefully, now you know all there is to know about Math.min! :)
I need to compare 4 values and determine which one is biggest.
OP asks for this implementation in a standard library:
int ensureRange(int value, int min, int max) {
return Math.min(Math.max(value, min), max);
}
boolean inRange(int value, int min, int max) {
return (value>= min) && (value<= max);
}
A pity the standard Math library lacks these
I understand this was asked for Java. In Android world, it's common to use Kotlin
and Java combined. In case some Kotlin user reached here (just like me), then they can use coerceIn extension function:
Kotlin Code:
println(10.coerceIn(1, 100)) // 10
println(10.coerceIn(1..100)) // 10
println(0.coerceIn(1, 100)) // 1
println(500.coerceIn(1, 100)) // 100
Read more on official Kotlin Documentation.