C，为什么我不能声明指针并初始化它？

Question

Well, I have the following code:

int main(int argc, char **argv, char **envp)
{
    const char *usuario= NULL;
    while(*envp)
    {
        char *str = *envp++;
        //if(strcmp(str,"USERNAME")==0)
        if(str[0] == 'U' && str[1] == 'S' && str[2]=='E' && str[3]=='R' && str[4] == 'N')
        {
            usuario = str;
            break;
        }
    }
    if(usuario != NULL)
    {
        printf("Hola, bienvenido al programa %s",usuario);
    }
    return 0;
}

And my question is why can't I declare the variable outside the while and initialize it inside?

char *str;
const char *usuario= NULL;
while(*envp)
{
    *str = *envp++;
    if(`...

The compiler says: warning: assignment makes integer from pointer without a cast

Answer 1

warning: assignment makes integer from pointer without a cast

The problem is that you're not assigning to your pointer, you're assigning to the value that the pointer points to:

*str = *envp++;

The * before str causes the pointer to be dereferenced. Instead, you probably want:

str = *envp++;

Answer 2

So the problem with this *str = *envp++; is that you are referencing the str pointer and then assigning a pointer to a char. In other words:

str is a pointer to a char

*envp is a pointer to a char

Thus

str = *envp++

would be a correct statement since both are pointers to a char. But in the provided code you are doing the following:

*str is a char

*envp is a pointer to a char

This means that you are trying to assign a "pointer to a char" to a char. The types do not match.

So fix this by changing your code like this:

char *str;
const char *usuario= NULL;
while(*envp)
{
    str = *envp++;
    if(... `