Thread: Segmentation fault when entering a for loop

I'm very new with C so bear with me.

Doing an assignment, I've run into a problem entering a for loop, where I get a segmentation fault.

Code:

for (i = 1; i < strlen(str); i++)

That's what the beginning of the for loop looks like, and it just never enters it. On the previous line, I print the value of strlen(str), which with the example I'm using is 2 (and gets printed as such), so it should be at least entering the loop.

I just don't understand why I would be getting a segmentation fault here. Maybe if I understood why I might be able to fix it, but I don't even understand why it happens, so that makes it a little harder to fix.

Any help would be greatly appreciated, and if more information is needed (short of posting the whole code, as I'm pretty sure that violates our academic conduct policy), just let me know.

There's nothing in that one line that could even begin to help us, if strlen(str) really gives 2 as you say it does.

What does the loop body look like? Perhaps the seg fault is in there and it just appears to be on that line. Seg faults can be tricky, and depending on how you "determined" where it was, the problem may actually lie elsewhere.

Posting the smallest, compilable code that produces the problem and giving us a test case to run would really help.

Also, learning to use a debugger would be a good idea. Then you can print all critical information at the exact time of the seg fault.

Yeah figured more would be needed, just wasn't sure exactly where to start.

Laptop battery is about to die though so I'll look into it and find some more relevant information when I get home.

Thanks.

Originally Posted by anduril462

Posting the smallest, compilable code that produces the problem and giving us a test case to run would really help.

OK, in the process of doing this right now, though I might just put it off until tomorrow because I'm exhausted right now. It doesn't appear to be this for loop itself though, as the loop by itself will compile and run properly. I guess this is where I'm getting confused. If it's not the for loop itself, I don't understand why it won't enter the loop.

Originally Posted by anduril462

What does the loop body look like? Perhaps the seg fault is in there and it just appears to be on that line. Seg faults can be tricky, and depending on how you "determined" where it was, the problem may actually lie elsewhere.

For what it's worth, I determined it to be at the beginning of the loop just by printing something immediately before the loop, and printing something immediately after entering the loop. The second print never happens, so I just assumed it was something wrong with the line posted above, but I guess that's not necessarily the case? Like I said, I'm extremely new to C, and I know next to nothing about segmentation faults, so I didn't really know where to start with this.

Thanks for the help so far, and thanks in advance for any future help.

Originally Posted by rehpot

For what it's worth, I determined it to be at the beginning of the loop just by printing something immediately before the loop, and printing something immediately after entering the loop. The second print never happens, so I just assumed it was something wrong with the line posted above, but I guess that's not necessarily the case? Like I said, I'm extremely new to C, and I know next to nothing about segmentation faults, so I didn't really know where to start with this.

So the when you use printf, it prints to stdout, which conventionally is buffered. You might print something, and before it makes it out of the output buffer and onto the screen, you get a seg fault. It will look like it never hit that printf statement, but it really did. Usually, it's line buffered, so as soon as it hits a new line, it flushes stdout and the data shows up on screen. It will also flush stdout if you ask for input on stdin. A better way is to make debug output go to stderr, which is unbuffered (you can use fprintf instead of printf), but the best is to learn to use a debugger.