strcpy给出分段错误

Question

Consider the following code:

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

char *cexpGen();
char *chartoStr(char c);

char c_exp[] = "1";
char expressions[2] = {'+', '-'};

int main()
{
    cexpGen();
    printf("%s", c_exp);
    return 0;
}

char *cexpGen()
{
    int now = 1;
    while ((c_exp[strlen(c_exp) - 1]) > 10)
    {
        if ((c_exp[strlen(c_exp) - 1]) == '+' || (c_exp[strlen(c_exp) - 1]) == '-')
        {
            strcpy(c_exp, chartoStr(((c_exp[strlen(c_exp) - 2]) + 1)));
            continue;
        }

        if (now = 1)
        {
            strcpy(c_exp, chartoStr(expressions[0]));
            now++;
            cexpGen();
        }

        if (now = 2)
        {
            strcpy(c_exp, chartoStr(expressions[1]));
            now++;
            cexpGen();
        }

        if (now = 3)
        {
            strcpy(c_exp, chartoStr(((c_exp[strlen(c_exp) - 1]) + 1)));
        }
    }
}

char *chartoStr(char c)
{
    char s[2] = {c, '\0'};
    return s;
}

I wanted to concatenate a cahracter and an string, But we don't have the function to do that, So I've defined a function chartoStr. Also c_exp and expressions varaibles aren't in read only mode, But strcpy() gives me segmentation fault. I also tried other functions any way, Like strcat, Which didn't help.

In case it helps, I debug it in VS Code. It opens strcpy-sse2-unaligned.S and shows segmentation fault in one of it's lines.

Is launch.json or task.json files needed? I don't think they may help, So I don't full the question with codes, But tell me if they're needed.

Answer 1

chartoStr() returns a local that is no longer available at function's end as well answered by @kiran Biradar

char s[2] = {c, '\0'};
return s;  // bad, UB

---

An alternative to calling chartoStr() it to create the string in the calling code using a compound literal (Since C99).

// strcpy(c_exp, chartoStr(expressions[0]));
//            v-----------------------------v---- compound literal         
strcpy(c_exp, (char []){expressions[0], '\0'});

A nice attribute is that there is no expensive allocation nor pointer to free. The compound literal is valid until the end of the block.

---

Note that code could employ other improvements by saving the length size_t len = strlen(c_exp) once early in cexpGen() and using it to append.

// strcpy(c_exp, chartoStr(expressions[0]));
c_exp[len++] = expressions[0];
c_exp[len] = '\0';

---

Other Problems

Take heed to @alk about char c_exp[] = "1";

while (c_exp[strlen(c_exp) - 1]) > 10 is undefined behavior should strlen(c_exp) return 0. Perhaps while ((len = strlen(c_exp)) > 0 && c_exp[len - 1]) > 10 ?

if (now = 1) is always true. @user3386109

Answer 2

You are getting segfault because you are returning address of local variable.

char s[2] = {c, '\0'};
    return s;

s will be destroyed once control exits chartoStr function.

Compiler is also warning the same

warning: function returns address of local variable [-Wreturn-local-addr]

  return s; ^ 

You don't need strcpy to copy the single character you can directly assign.

To solve your problem you can try as below.

char *chartoStr(char c)
{
    char *s = malloc(2);
     s[0] = c;
     s[1] =  '\0';
    return s;
}

Don't forget to free s after work done.