Question

所以我有一段代码，我有一个数组（ar = [1,2,3,4,5]）和两个函数。这些函数应该完全相同：如果数组中存在1，则打印出一些东西。但是函数func总是返回＆＃34; nope＆＃34;而不是＆＃34;一个＆＃34;但功能另一个总是返回正确的东西。两个函数之间的唯一区别是函数func有一个开关而不是if / else。为什么？在源代码中有大约12个案例，所以我实际上需要使用开关。

var ar=[1,2,3,4,5];
function func(num){
    var one=num;
    switch (one) {
        case one===1:
            console.log("one");
            break;
        default:
            console.log("nope");
            break;
    }
}
function another (num) {
     if(num===2){
        console.log("found two");
     } else if(num===3){
        console.log("found thre");
     } else{
        console.log("nope");
     }
}

ar.forEach(func);
ar.forEach(another);

Answer 1

您必须使用要与一个

进行比较的值

因此

而不是

case 1:

这是一个小提琴 https://jsfiddle.net/cunx1ono/

Answer 2

最简单的方法。如果要在大小写中使用比较，请将开关参数更改为public void onSurfaceCreated(GL10 glUnused, EGLConfig config) { ... final float eyeX = 0.0f; final float eyeY = 0.0f; final float eyeZ = 7.0f; final float lookX = 0.0f; final float lookY = 0.0f; final float lookZ = 0.0f; final float upX = 0.0f; final float upY = 1.0f; final float upZ = 0.0f; Matrix.setLookAtM(mViewMatrix, 0, eyeX, eyeY, eyeZ, lookX, lookY, lookZ, upX, upY, upZ); ... } public void onSurfaceChanged(GL10 glUnused, int width, int height) { ... final float ratio = width / height; final float left = -ratio; final float right = ratio; final float bottom = -1.0f; final float top = 1.0f; final float near = 1.0f; final float far = 20.0f; Matrix.frustumM(mProjectionMatrix, 0, left, right, bottom, top, near, far); ... } public void onDrawFrame(GL10 glUnused) { ... Matrix.multiplyMM(mMVPMatrix, 0, mViewMatrix, 0, mModelMatrix, 0); GLES20.glUniformMatrix4fv(mMVMatrixHandle, 1, false, mMVPMatrix, 0); Matrix.multiplyMM(mMVPMatrix, 0, mProjectionMatrix, 0, mMVPMatrix, 0); GLES20.glUniformMatrix4fv(mMVPMatrixHandle, 1, false, mMVPMatrix, 0); ... }，因为true返回true / false。这就是为什么你总是得到＆＃34; nope＆＃34;。

one===1

Switch语句表现得不像它应该的那样

2 个答案: