从分配函数返回一个指针 [英] Assigning a pointer returned from a function

问题描述

我有一个程序，它看起来像这样：

I have a program which looks like this:

EVP_PKEY *generate_RSA_key_and_uuid(unsigned char uuid[16])
{
    EVP_PKEY        *key_p;
    key_p = EVP_PKEY_new();
    return key_p;  
}

int build_and_save_csr(int dn_entries, X509_REQ *req_p, EVP_PKEY *priv_key_p, char *passphrase)
{   
    priv_key_p = generate_RSA_key_and_uuid(uuid);
}

int makecsr(X509_REQ *req_p, EVP_PKEY *priv_key_p, char *passphrase)
{

    if (build_and_save_csr(2, req_p, priv_key_p, passphrase) != 0) {
        fprintf(stderr, "Could not create csr file / private key...\n");
        exit(1);
    }

    return 0;
}

int main(int argc, char *argv[])
{

    EVP_PKEY        *priv_key_p;
    X509_REQ        *req_p;

    makecsr(req_p, priv_key_p, passphrase);
    return 0;
}

在此已经用完， * priv_key_p 应包含 * generate_RSA_key_and_uuid 返回的值。

After this has run, *priv_key_p should contain the value returned from *generate_RSA_key_and_uuid.

在gdb的，我可以看到，在 generate_RSA_and_uuid 的 key_p 的末尾包含正确的值，但在 priv_key_p = generate_RSA_key_and_uuid（UUID）; 已执行， priv_key_p 为空

In gdb, I can see that at the end of generate_RSA_and_uuid that key_p contains the correct value but after priv_key_p = generate_RSA_key_and_uuid(uuid); has executed, priv_key_p is empty.

有谁知道如何正确值分配给指针？

Does anybody know how to properly assign the value to the pointer?

推荐答案

您需要额外的间接水平，否则你只是修改指针的本地副本没有东西返回：

You need an extra level of indirection, otherwise you're just modifying local copies of the pointers and nothing gets returned:

int build_and_save_csr(int dn_entries, X509_REQ *req_p, EVP_PKEY **priv_key_p, char *passphrase)
{   
    *priv_key_p = generate_RSA_key_and_uuid(uuid);
}

int makecsr(X509_REQ *req_p, EVP_PKEY **priv_key_p, char *passphrase)
{
    if (build_and_save_csr(2, req_p, priv_key_p, passphrase) != 0) {
        fprintf(stderr, "Could not create csr file / private key...\n");
        exit(1);
    }
    return 0;
}

int main(int argc, char *argv[])
{
    EVP_PKEY        *priv_key_p;
    X509_REQ        *req_p;

    makecsr(req_p, &priv_key_p, passphrase);
    return 0;
}

另外，您可以使用函数的结果（如你在做已经 generate_RSA_key_and_uuid（））返回指针（也许返回出错NULL） - 这个出手的需要间接的额外级别。

Alternatively you can use the function result (as you do already in generate_RSA_key_and_uuid()) to return the pointer (perhaps return NULL on error) - this gets rid of the need for the extra level of indirection.

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