How to use signal name as variable name in generated code
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Andrea Carignano
il 17 Dic 2025 alle 22:37
Modificato: Walter Roberson
il 20 Dic 2025 alle 8:49
I have a simple model with an atomic subsystem block.
the signal after the subsystem "Function" is called "x" in the model.
The generated code for the step function is
void untitled1_step(void)
{
real32_T rtb_Gain;
/* Outputs for Atomic SubSystem: '<Root>/Function' */
/* Inport: '<Root>/In1' */
rtb_Gain = untitled1_Function(untitled1_U.In1);
/* End of Outputs for SubSystem: '<Root>/Function' */
/* Outport: '<Root>/Out2' incorporates:
* Bias: '<Root>/Bias'
*/
untitled1_Y.Out2 = rtb_Gain + 2.0F;
}
Embedded coder generate rtb_Gain variable.
However I would like to generate the x variable instead, so that the code became:
void untitled1_step(void)
{
real32_T rtb_Gain;
/* Outputs for Atomic SubSystem: '<Root>/Function' */
/* Inport: '<Root>/In1' */
x = untitled1_Function(untitled1_U.In1);
/* End of Outputs for SubSystem: '<Root>/Function' */
/* Outport: '<Root>/Out2' incorporates:
* Bias: '<Root>/Bias'
*/
untitled1_Y.Out2 = x + 2.0F;
}
Is there any general setting that I can activate to do this?
2 Commenti
Walter Roberson
il 20 Dic 2025 alle 4:38
If you assign to x, why are you later using rtb_Gain + 2.0F instead of x + 2.0F ?
Risposte (1)
Fangjun Jiang
il 19 Dic 2025 alle 19:34
Double click the signal line and set the signal name as 'x'. That might be sufficient for your need.
Currently, the 'x' shown in your model looks like an annotation, not the signal name.
For more control, select the signal line, right click, select "Property" and set the signal line property.
One of the ways is to check "Name must resolve to signal object" and create a Simulink.Signal object in base workspace. Check the document for Simulink.Signal to learn more.
x=Simulink.Signal
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