SIGSEGV内存访问冲突

Question

I am writing a school project in C and apparently something in this code is causing it to crash with SIGSEGV memory access violation error even though it compiles without any problems.

int P, H, S, R;
char *end;

if (argc != 5){
    printf("Incorrect ammount of parameters.\n");
    exit(1);
}

errno = 0;
P=strtod(argv[2],&end);
H=strtod(argv[3],&end);
S=strtod(argv[4],&end);
R=strtod(argv[5],&end);

if (errno != 0){
    printf("Wrong input: %s\n",strerror(errno));
    exit(1);
}

//I know have only integers in P,H,S,R, that's why I can afford to do the following. Please don't judge me.
if ((H < 0) || (S < 0) || (R < 0) || (H > 5000) || ( S > 5000) || ( R > 5000)){
    printf("Incorrect waiting time H,S,R >= 0 && H,S,R < 5001\n");
    exit(1);
}

if ((P < 1) || ((P % 2)==1)){
    printf("Must be even number bigger than 0.");
    exit(1);
}

Any ideas? EDIT: even if I put printf at the very beginning of the code i does not print anything, only memory access violation error which looks like this

Neoprávněný přístup do paměti (SIGSEGV) (core dumped [obraz paměti uložen])

Answer 1

It would be best if you could track down the exact line where the error is occurring. You can do that in a debugger, or simply by putting printf statements all over the place and seeing what the last one that works is.

However, from a quick look, I think that the problem is here:

R=strtod(argv[5],&end);

SIGSEGV (segmentation fault) means that you attempted an invalid memory access. A few lines up, you checked that argc == 5 . This means that the valid indices for argv are 0 through 4. There is no argv[5] , and any attempt to read it results in undefined behaviour.

Answer 2

The problem is here:

if (argc != 5)
…
R=strtod(argv[5],&end);

When argc == 5 , argv[5] == NULL . You crash when you convert a null pointer. You should be using strtod() on argv[1] through argv[4] . It is unusual to use strtod() to convert strings to integers; usually, you'd use strtol() or one of its relatives.