如果你想在屏幕上可视化一个值的变化，你可以使用类似于这个的热图函数(我用hlsl写的，但很容易适应glsl):

float4 HeatMapColor(float value, float minValue, float maxValue)
{
    #define HEATMAP_COLORS_COUNT 6
    float4 colors[HEATMAP_COLORS_COUNT] =
    {
        float4(0.32, 0.00, 0.32, 1.00),
        float4(0.00, 0.00, 1.00, 1.00),
        float4(0.00, 1.00, 0.00, 1.00),
        float4(1.00, 1.00, 0.00, 1.00),
        float4(1.00, 0.60, 0.00, 1.00),
        float4(1.00, 0.00, 0.00, 1.00),
    };
    float ratio=(HEATMAP_COLORS_COUNT-1.0)*saturate((value-minValue)/(maxValue-minValue));
    float indexMin=floor(ratio);
    float indexMax=min(indexMin+1,HEATMAP_COLORS_COUNT-1);
    return lerp(colors[indexMin], colors[indexMax], ratio-indexMin);
}

然后在像素着色器中输出如下内容:

return HeatMapColor(myValue, 0.00, 50.00);

并且可以了解它在像素之间的变化情况:

当然你可以使用任何你喜欢的颜色。

现有的答案都是好东西，但我想分享更多的小宝石，在调试棘手的精度问题在GLSL着色器有价值。对于以浮点数表示的非常大的int数，需要注意正确使用floor(n)和floor(n + 0.5)来实现round()为精确的int。然后，可以通过以下逻辑将字节组件打包到R、G和B输出值中，呈现一个精确int的浮点值。

  // Break components out of 24 bit float with rounded int value
  // scaledWOB = (offset >> 8) & 0xFFFF
  float scaledWOB = floor(offset / 256.0);
  // c2 = (scaledWOB >> 8) & 0xFF
  float c2 = floor(scaledWOB / 256.0);
  // c0 = offset - (scaledWOB << 8)
  float c0 = offset - floor(scaledWOB * 256.0);
  // c1 = scaledWOB - (c2 << 8)
  float c1 = scaledWOB - floor(c2 * 256.0);

  // Normalize to byte range
  vec4 pix;  
  pix.r = c0 / 255.0;
  pix.g = c1 / 255.0;
  pix.b = c2 / 255.0;
  pix.a = 1.0;
  gl_FragColor = pix;

void main(){
  float bug=0.0;
  vec3 tile=texture2D(colMap, coords.st).xyz;
  vec4 col=vec4(tile, 1.0);

  if(something) bug=1.0;

  col.x+=bug;

  gl_FragColor=col;
}

你可以试试这个:https://github.com/msqrt/shader-printf，它是一个叫做“GLSL的简单打印功能”的实现。

You might also want to try ShaderToy, and maybe watch a video like this one (https://youtu.be/EBrAdahFtuo) from "The Art of Code" YouTube channel where you can see some of the techniques that work well for debugging and visualising. I can strongly recommend his channel as he writes some really good stuff and he also has a knack for presenting complex ideas in novel, highly engaging and and easy to digest formats (His Mandelbrot video is a superb example of exactly that : https://youtu.be/6IWXkV82oyY)

我希望没有人介意这个迟到的回复，但这个问题在谷歌搜索GLSL调试中排名很高，当然在9年里发生了很大变化:-)

PS:其他替代方案也可以是NVIDIA nSight和AMD ShaderAnalyzer，它们为着色器提供了一个完整的步进调试器。

我发现变换反馈是调试顶点着色器的有用工具。您可以使用它来捕获VS输出的值，并在CPU端读取它们，而不必通过光栅化器。

这里是另一个转换反馈教程的链接。

在这个答案的底部是一个GLSL代码的例子，它允许输出完整的浮点值作为颜色，编码IEEE 754 binary32。我像这样使用它(这段代码给出了modelview矩阵的yy组件):

vec4 xAsColor=toColor(gl_ModelViewMatrix[1][1]);
if(bool(1)) // put 0 here to get lowest byte instead of three highest
    gl_FrontColor=vec4(xAsColor.rgb,1);
else
    gl_FrontColor=vec4(xAsColor.a,0,0,1);

在屏幕上显示后，您可以选择任何颜色选择器，将颜色格式化为HTML(如果您不需要更高的精度，则将00附加到rgb值，如果需要，则执行第二遍以获得较低的字节)，然后您将得到浮点数的十六进制表示为IEEE 754 binary32。

下面是toColor()的实际实现:

const int emax=127;
// Input: x>=0
// Output: base 2 exponent of x if (x!=0 && !isnan(x) && !isinf(x))
//         -emax if x==0
//         emax+1 otherwise
int floorLog2(float x)
{
    if(x==0.) return -emax;
    // NOTE: there exist values of x, for which floor(log2(x)) will give wrong
    // (off by one) result as compared to the one calculated with infinite precision.
    // Thus we do it in a brute-force way.
    for(int e=emax;e>=1-emax;--e)
        if(x>=exp2(float(e))) return e;
    // If we are here, x must be infinity or NaN
    return emax+1;
}

// Input: any x
// Output: IEEE 754 biased exponent with bias=emax
int biasedExp(float x) { return emax+floorLog2(abs(x)); }

// Input: any x such that (!isnan(x) && !isinf(x))
// Output: significand AKA mantissa of x if !isnan(x) && !isinf(x)
//         undefined otherwise
float significand(float x)
{
    // converting int to float so that exp2(genType) gets correctly-typed value
    float expo=float(floorLog2(abs(x)));
    return abs(x)/exp2(expo);
}

// Input: x\in[0,1)
//        N>=0
// Output: Nth byte as counted from the highest byte in the fraction
int part(float x,int N)
{
    // All comments about exactness here assume that underflow and overflow don't occur
    const float byteShift=256.;
    // Multiplication is exact since it's just an increase of exponent by 8
    for(int n=0;n<N;++n)
        x*=byteShift;

    // Cut higher bits away.
    // $q \in [0,1) \cap \mathbb Q'.$
    float q=fract(x);

    // Shift and cut lower bits away. Cutting lower bits prevents potentially unexpected
    // results of rounding by the GPU later in the pipeline when transforming to TrueColor
    // the resulting subpixel value.
    // $c \in [0,255] \cap \mathbb Z.$
    // Multiplication is exact since it's just and increase of exponent by 8
    float c=floor(byteShift*q);
    return int(c);
}

// Input: any x acceptable to significand()
// Output: significand of x split to (8,8,8)-bit data vector
ivec3 significandAsIVec3(float x)
{
    ivec3 result;
    float sig=significand(x)/2.; // shift all bits to fractional part
    result.x=part(sig,0);
    result.y=part(sig,1);
    result.z=part(sig,2);
    return result;
}

// Input: any x such that !isnan(x)
// Output: IEEE 754 defined binary32 number, packed as ivec4(byte3,byte2,byte1,byte0)
ivec4 packIEEE754binary32(float x)
{
    int e = biasedExp(x);
    // sign to bit 7
    int s = x<0. ? 128 : 0;

    ivec4 binary32;
    binary32.yzw=significandAsIVec3(x);
    // clear the implicit integer bit of significand
    if(binary32.y>=128) binary32.y-=128;
    // put lowest bit of exponent into its position, replacing just cleared integer bit
    binary32.y+=128*int(mod(float(e),2.));
    // prepare high bits of exponent for fitting into their positions
    e/=2;
    // pack highest byte
    binary32.x=e+s;

    return binary32;
}

vec4 toColor(float x)
{
    ivec4 binary32=packIEEE754binary32(x);
    // Transform color components to [0,1] range.
    // Division is inexact, but works reliably for all integers from 0 to 255 if
    // the transformation to TrueColor by GPU uses rounding to nearest or upwards.
    // The result will be multiplied by 255 back when transformed
    // to TrueColor subpixel value by OpenGL.
    return vec4(binary32)/255.;
}