为什么第二个for循环不执行？

以下代码中的问题是第二个for循环似乎未执行。

我添加了printf("1");andprintf("h");语句以测试循环是否正在执行。程序打印1，但不打印h。

有关该程序的一些背景信息是，我应该分配一个2D数组，并在该数组的每一行中以不同的频率存储正弦波。

变量PI也定义为带有值的宏acos(-1).

short int** Notes = (short int**) calloc(25,sizeof(short int*));
for(int x = 0; x < 25; x++){
    Notes[x] = (short int*) calloc(44100 * (1/3),sizeof(short int));
    }

    for(int x = 0; x < 25; x++){
        float freq;
        switch(x){
            case 0:
                freq = 440.00;
                break;
            case 1:
                freq = 466.16;
                break;
            case 2:
                freq = 493.88;
                break;
            case 3:
                freq = 523.25;
                break;
            case 4:
                freq = 554.37;
                break;
            case 5:
                freq = 587.33;
                break;
            case 6:
                freq = 622.25;
                break;
            case 7:
                freq = 659.25;
                break;
            case 8:
                freq = 698.46;
                break;
            case 9:
                freq = 739.99;
                break;
            case 10:
                freq = 783.99;
                break;
            case 11:
                freq = 830.61;
                break;
            case 12:
                freq = 880.00;
                break;
            case 13:
                freq = 932.33;
                break;
            case 14:
                freq = 987.77;
                break;
            case 15:
                freq = 1046.50;
                break;
            case 16:
                freq = 1109.73;
                break;
            case 17:
                freq = 1174.66;
                break;
            case 18:
                freq = 1244.51;
                break;
            case 19:
                freq = 1318.51;
                break;
            case 20:
                freq = 1396.91;
                break;
            case 21:
                freq = 1479.98;
                break;
            case 22:
                freq = 1567.98;
                break;
            case 23:
                freq = 1661.22;
                break;
            case 24:
                freq = 0.00;
                break;                  
        }
        printf("1");
        for(int y = 0; y < 44100 * (1/3); y++)
        {
            Notes[x][y] = 32700 * sin(2 * PI * freq * (y/44100));
            printf("h");
        }
    }