Error:Assignment makes pointer from integer without a cast

Question

int findNumber(char *exp,int i,int *num)    
{    
    int k=i;    
    char *p;    
    p=exp[i]; //<-- here
    while(*p>='0'&&*p<='9')    
    {
        (*num)=(*num)*10+(*p);
        k++;
        p++;
    }
    return k;
}

i keep getting that error in line: (p=exp[i];) Im trying to send a char array, and (i,num) integers, the 'i' im just putting it to be 0 for now, until the code works so dont give attention to it. but the function should return the place of the first character in "exp" that is not a number, with being sure that all the ones before are numbers.

simonc · Accepted Answer · 2013-01-09 19:33:26Z

5

p is a char* so you need to assign a pointer to it but exp[i] returns a single char element from an array. Try

p = &exp[i];

or

p = (exp + i);

instead.

edited Jan 9, 2013 at 19:33

answered Jan 9, 2013 at 18:30

simonc

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Comments

hmatar · Accepted Answer · 2013-01-09 18:35:56Z

0

do correct the line to correspond to the following p=&(exp[i]); this means that you are assigning the pointer of exp[i] to p.

answered Jan 9, 2013 at 18:35

hmatar

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Comments

Jack · Accepted Answer · 2013-01-09 19:13:15Z

0

You need take the address-of instead of and then pass to p, that's a pointer. In other words,you are assign char to a char* returned from value atiindex inexp` array.

Try this: p = &exp[i];

answered Jan 9, 2013 at 19:13

Jack

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