中国DOS联盟

```
echo off
set a=0
for /l %%i in (71,1,99) do set /a a+=1 & echo %a%

The %a% here always shows 0. It seems that set /a a+=1 doesn't work at all. What's the reason for this? Is it that set /a can't be used in for /l?
```

LZ didn't understand the usage

@echo off
set a=0
for /l %%i in (71,1,99) do set /a a+=1
echo %a%
pause

If you want to use it on the same line, you must enable delayed variables (specifically search the forum)

@echo off
set a=0
for /l %%i in (71,1,99) do set /a a+=1 & call echo %%a%%

pause>nul

I also have a place I don't understand. What does "set /a a+=1" mean? The key is what the usage of a+=1 is. Which friend who knows tells me, thank you.

The `set /a a+=1` is equivalent to `set /a a=%a%+1` in the context of batch scripting in Windows. So the translation is: `set /a a+=1` means equal to `set /a a=%a%+1`

Thank you, friend from the upper floor.

Originally posted by oilio at 2007-3-10 12:54:
I also have a question I don't understand. What does "set /a a+=1" mean? The key is what the usage of a+=1 is for? Friends who know, please tell me, thank you.

Friends who have learned other programming languages may know this usage at a glance. There are many similar usages, such as: set /a a*=2, set /a a-=5, etc., which respectively mean set /a a=a*2, set /a a=a-5. Generally, set /a a+=1 is called an accumulator. Each time it runs, the value of a increases by 1, which is mainly used to control the program flow.

Thanks, Brother lxmxn. I really don't know what to say, thanks! This is the Nth time I've seen this usage. Because I used to be very weak, now my level is a little better than before. Today, when I saw this usage again, I thought it was time, and couldn't help but want to ask. Thanks everyone, especially Brother lxmxn for listing several examples. Brother lxmxn mentioned the accumulator. I think that in this batch processing, using set /a a+=1, the main function is to calculate that from 71 with a step value of 1 increasing to 99, a total of 28 increments occur. But to get the total number of numbers, the number of times does not include 71 itself, so when calculating the total number of numbers, one is less. So here set /a a+=1 is used. I don't know if my understanding is correct? It seems that I made a gain today, and I understand two things at once. I'll practice it to deepen my impression. Thanks everyone again.

Last edited by oilio on 2007-3-10 at 09:14 PM ]

Originally posted by oilio at 2007-3-10 20:21:
Thank you, brother lxmxn. I really don't know what to say, thank you! This is the Nth time I've seen this usage. Because I've been very weak before, and now my level is a little better than before. Today, when I see this usage again, I feel that...

Brother, there's no need to be so polite. Everyone in the forum is also a fate. Let's learn and progress together.

I'm really happy to see that you've made great progress these days.

Originally posted by oilio at 2007-3-10 20:21:
I think that the main function of using set /a a+=1 in this batch processing is to calculate that from step size 1 increasing from 77 to 99, a total of 28 increments occur, but to get the total number of numbers, the count does not include 71 itself, so when calculating the total number of numbers, it will be one less. So here set /a a+=1 is used. I don't know if my understanding is correct?

Brother's understanding is a bit deviated. Here, there are a total of 99-71+1=29 increments, because the for /l loop includes the first number, which is 71. You can try for /l %a in (1,1,29) do, and the result should be the same. The function of set /a a+=1 is not to increase the increment by one, but to make the value of a increase by 1 each time the loop is executed. Here, the loop is executed 29 times, so the final result should be 29.

It seems incorrect. Let me think. If it's for /l %a in (1,1,3), from 1 to 2 is an increase of 1 time, and from 2 to 3 is the second increase. There are a total of 2 times. But there are actually three numbers here. Shouldn't the number of increments be 2 times? This command should be calculating the number from 1 to 3. I don't know if my understanding is wrong, heh heh. Sometimes I feel quite smart, but sometimes I get stuck, heh heh. Tired, going to sleep. I'll think about it again. Maybe I can figure it out after resting, heh heh. Thanks to brother lxxm for the reply, which has benefited me a lot.

for /l %a in (1,1,3) do
When %a=1, count once
When %a=2, count once
When %a=3, count once
So it should be 3 times

Thanks, Brother Zhou Shijie. So my problem was here. The key is that I mentioned incrementing and forgot about this. If I start counting from the first number, that's right. I confused myself. Haha. Thanks.

I tried many times before, and I couldn't use it on the same line. It turns out that using variable delay can do it. How exactly to use it?

Here are two ways to use variable delay:

1. Enable variable delay:

@echo off
setlocal EnableDelayedExpansion
set a=0
for /l %%i in (71,1,99) do set /a a+=1&echo !a!
pause>nul

2. Use call

@echo off
set a=0
for /l %%i in (71,1,99) do set /a a+=1&call echo %%a%%
pause>nul