为什么0 + 1 == 49？

Question

我是MatLab的超级初学者,我正在尝试调试一个我正在编写的简单脚本.我在尝试调试代码时遇到了一个奇怪的错误.这是脚本:

function [prob] = QuantumHW1(j,k,m)

X = [0 1; 1 0];
Y = [0 -sqrt(-1); sqrt(-1) 0];
Z = [1 0; 0 -1];
H = 1/sqrt(2) * [1 1; 1 -1];
S = [1 0; 0 i];
T = [1 0; 0 exp(sqrt(-1)*pi/4)];

mats = {X,Y,Z,H,S,T};

binJ = dec2bin(j,k);
binM = dec2bin(m,k);

totOps = {};

%Set up all the operators to be used
for p = 1:k
    totOps(p) = mats(mod(p,6));
    if p == 0
        totOps(p) = X;
    end
end

withM = {};

%Dot product with M
for p = 1:k
    p
    binM(p)+1
    totOps(:,1)
    withM(p) = totOps(:,binM(p)+1);
end

rTotal = 0;

%Now take components with respect to J
for p = 1:k
    rTotal = rTotal + [not(binJ(p)),binJ(p)] * withM(p);
end

prob = norm(runningTotal)^2;

disp('The probability to measure j = %d in a k = %d system on input m = %d is %d',j,k,m,prob);
end

当我运行程序时,我在行上得到一个Array Index Out of Bounds错误withM(p) = totOps(:,binM(p)+1);.我试着确保p的值是正确的.在通过for循环的第一次迭代中,binM(p)= 0.但是当我尝试获得binM(p)+ 1时,我得到49.这非常奇怪.

任何帮助深表感谢.我正试图弄清楚为什么会发生这种情况.

Answer 1

因为binM(p)保持字符串'0'的ASCII值,而不是实际的double值0.并且'0'的ASCII值是48.'0'+ 1自动转换为double值.你做剩下的数学运算.