# Documentation

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## Troubleshoot Variables in `parfor`-Loops

### Ensure That `parfor`-Loop Variables Are Consecutive Increasing Integers

Loop variables in a `parfor`-loop must be consecutive increasing integers. For this reason, the following examples return errors:

```parfor i = 0:0.2:1 % not integers parfor j = 1:2:11 % not consecutive parfor k = 12:-1:1 % not increasing```
You can fix these errors by converting the loop variables into a valid range. For example, you can fix the noninteger example as follows:
```iValues = 0:0.2:1; parfor idx = 1:numel(iValues) i = iValues(idx); ... end```

### Solve Variable Classification Issues in `parfor`-Loops

When MATLAB® recognizes a name in a `parfor`-loop as a variable, the variable is classified in one of several categories, shown in the following table. Make sure that your variables are uniquely classified and meet the category requirements. `parfor`-loops that violate the requirement return an error.

ClassificationDescription
Loop VariablesLoop indices
Sliced VariablesArrays whose segments are operated on by different iterations of the loop
Broadcast VariablesVariables defined before the loop whose value is required inside the loop, but never assigned inside the loop
Reduction VariablesVariables that accumulates a value across iterations of the loop, regardless of iteration order
Temporary VariablesVariables created inside the loop, and not accessed outside the loop

To find out which variables you have, examine the code fragment. All variable classifications in the table are represented in this code:

If you run into variable classification problems, consider these approaches before you resort to the more difficult method of converting the body of a `parfor`-loop into a function.

• If you use a nested `for`-loop to index into a sliced array, you cannot use that array elsewhere in the `parfor`-loop. The code on the left does not work because `A` is sliced and indexed inside the nested `for`-loop. The code on the right works because `v` is assigned to `A` outside the nested loop. You can compute an entire row, and then perform a single assignment into the sliced output.

InvalidValid
```A = zeros(4, 10); parfor i = 1:4 for j = 1:10 A(i, j) = i + j; end disp(A(i, 1)) end```
```A = zeros(4, 10); parfor i = 1:4 v = zeros(1, 10); for j = 1:10 v(j) = i + j; end disp(v(1)) A(i, :) = v; end```

• The code on the left does not work because the variable `x` in `parfor` cannot be classified. This variable cannot be classified because there are multiple assignments to different parts of `x`. Therefore `parfor` cannot determine whether there is a dependency between iterations of the loop. The code on the right works because you completely overwrite the value of `x`. `parfor` can now determine unambiguously that `x` is a temporary variable.

InvalidValid
```parfor idx = 1:10 x(1) = 7; x(2) = 8; out(idx) = sum(x); end```
```parfor idx = 1:10 x = [7, 8]; out(idx) = sum(x); end```

• This example shows how to slice the field of a structured array. See `struct` for details. The code on the left does not work because the variable `a` in `parfor` cannot be classified. This variable cannot be classified because the form of indexing is not valid for a sliced variable. The first level of indexing is not the sliced indexing operation, even though the field `x` of `a` appears to be sliced correctly. The code on the right works because you extract the field of the `struct` into a separate variable `tmpx`. `parfor` can now determine correctly that this variable is sliced. In general, you cannot use fields of `struct`s or properties of objects as sliced input or output variables in `parfor`.

InvalidValid
```a.x = []; parfor idx = 1:10 a.x(idx) = 7; end```
```tmpx = []; parfor idx = 1:10 tmpx(idx) = 7; end a.x = tmpx;```

### Structure Arrays in parfor-Loops

#### Creating Structures as Temporaries

You cannot create a structure in a `parfor`-loop using dot notation assignment. In the code on the left, both lines inside the loop generate a classification error. In the code on the right, as a workaround you can use the `struct` function to create the structure in the loop or in the first field.

InvalidValid
```parfor i = 1:4 temp.myfield1 = rand(); temp.myfield2 = i; end ```
```parfor i = 1:4 temp = struct(); temp.myfield1 = rand(); temp.myfield2 = i; end```
```parfor i = 1:4 temp = struct('myfield1',rand(),'myfield2',i); end```

#### Slicing Structure Fields

You cannot use structure fields as sliced input or output arrays in a `parfor`-loop. In other words, you cannot use the loop variable to index the elements of a structure field. In the code on the left, both lines in the loop generate a classification error because of the indexing. In the code on the right, as a workaround for sliced output, you employ separate sliced arrays in the loop. Then you assign the structure fields after the loop is complete.

InvalidValid
```parfor i = 1:4 outputData.outArray1(i) = 1/i; outputData.outArray2(i) = i^2; end ```
```parfor i = 1:4 outArray1(i) = 1/i; outArray2(i) = i^2; end outputData = struct('outArray1',outArray1,'outArray2',outArray2); ```

The workaround for sliced input is to assign the structure field to a separate array before the loop. You can use that new array for the sliced input.

```inArray1 = inputData.inArray1; inArray2 = inputData.inArray2; parfor i = 1:4 temp1 = inArray1(i); temp2 = inArray2(i); end```

### Converting the Body of a `parfor`-Loop into a Function

If all else fails, you can usually solve variable classification problems in `parfor`-loops by converting the body of the `parfor`-loop into a function. In the code on the left, Code Analyzer flags a problem with variable y, but cannot resolve it. In the code on the right, you solve this problem by converting the body of the `parfor`-loop into a function.

InvalidValid
```function parfor_loop_body_bad data = rand(5,5); means = zeros(1,5); parfor X = 1:5 % Code Analyzer flags problem % with variable y below y.mean = mean(data(:,X)); means(X) = y.mean; end disp(means); ```
```function parfor_loop_body_good data = rand(5,5); means = zeros(1,5); parfor X = 1:5 % Call a function instead means(X) = computeMeans(data(:,X); end disp(means); % This function now contains the body % of the parfor-loop function meansX = computeMeans(dataX) y.mean = mean(dataX); meansX = y.mean;```
```Starting parallel pool (parpool) using the 'local' profile ... connected to 4 workers. 0.6786 0.5691 0.6742 0.6462 0.6307```

### Unambiguous Variable Names

If you use a name that MATLAB cannot unambiguously distinguish as a variable inside a `parfor`-loop, at parse time MATLAB assumes you are referencing a function. Then at run-time, if the function cannot be found, MATLAB generates an error. See Variable Names (MATLAB). For example, in the following code `f(5)` could refer either to the fifth element of an array named `f`, or to a function named `f` with an argument of `5`. If `f` is not clearly defined as a variable in the code, MATLAB looks for the function `f` on the path when the code runs.

```parfor i = 1:n ... a = f(5); ... end```

### Transparent `parfor`-loops

The body of a `parfor`-loop must be transparent: all references to variables must be “visible” in the text of the code. For more details about transparency, see Ensure Transparency in parfor-Loops.

### Global and Persistent Variables

The body of a `parfor`-loop cannot contain `global` or `persistent` variable declarations.