C++中调用resize()函数时出现“No instance of overloaded function”错误的解决方法咨询

Hey there! Let's break down what's going wrong with your code and get that IDFT function working properly.

Why You're Seeing That Error #

The line causing the overload mismatch is this one:

x.resize(Xf.size(), static_cast<std::vector<float>>(0, 0));

Here's the issue:

• The std::vector::resize method's second parameter expects an element of the vector's type (in your case, a float), not another std::vector<float>.
• Your static_cast attempt is invalid syntax for creating a vector anyway—you don't need to construct a vector here at all.

The Correct Way to Resize and Initialize #

To resize x to match the size of Xf and set all elements to 0, you just need to pass the desired size and a float value of 0. Either of these will work:

// Explicitly set initial value to 0.0f
x.resize(Xf.size(), 0.0f);

// Or omit the second parameter—float elements default to 0
x.resize(Xf.size());

Fixing Other IDFT Logic Bugs #

While we're at it, there are a few logic issues in your loops that will prevent the IDFT from calculating correctly:

1. You're using x[k] instead of Xf[k]: Since x starts at 0, multiplying it by the exponential term will always leave it at 0. You need to use the input values from Xf.
2. Wrong exponential sign: Standard IDFT uses a negative exponent (opposite of DFT).
3. Missing normalization: IDFT requires dividing the final result by the vector size to get the correct amplitude.

Here's the corrected full function:

#include <vector>
#include <complex>
#include <cmath>

// Define PI if it's not already available
#ifndef PI
#define PI 3.14159265358979323846
#endif

void IDFT(const std::vector<std::complex<float>> &Xf, std::vector<float> &x) {
    // Resize x to match Xf's size and initialize to 0
    x.resize(Xf.size(), 0.0f);

    const size_t N = Xf.size();
    for (size_t k = 0; k < N; ++k) {
        for (size_t m = 0; m < N; ++m) {
            // Negative exponent for IDFT calculation
            std::complex<float> expval(0, -2 * PI * static_cast<float>(k * m) / N);
            // Accumulate the real part of Xf[k] * exp into x[m]
            x[m] += std::real(Xf[k] * std::exp(expval));
        }
    }

    // Normalize the result by the vector size
    for (auto &val : x) {
        val /= N;
    }
}

Quick Recap #

• Use x.resize(Xf.size(), 0.0f) (or the shorter version without the second parameter) to fix the overload error.
• Adjust the IDFT logic to use input values from Xf, a negative exponent, and normalize the final result.

内容的提问来源于stack exchange，提问作者LYM