The meaning of -e option in the Intel SGX enclave linker flag

To create an SGX enclave, we have to pass long ldflags:

        -Wl,--whole-archive -l$(Trts_Library_Name) -Wl,--no-whole-archive \
        -Wl,--whole-archive -lsgx_tcmalloc -Wl,--no-whole-archive \
        -Wl,--start-group -lsgx_tstdc -lsgx_tcxx -l$(Crypto_Library_Name) -l$(Service_Library_Name) -Wl,--end-group \
        -Wl,-Bstatic -Wl,-Bsymbolic -Wl,--no-undefined \
        -Wl,-pie,-eenclave_entry -Wl,--export-dynamic  \
        -Wl,--defsym,__ImageBase=0 \
        -Wl,--version-script=Enclave/Enclave.lds

All the other options are somewhat understood, but I don't get what does the -pie,-eenclave_entry do.

To my understanding, enclave_entry is an assembly routine defined in sdk/trts/linux/trts_pic.S

DECLARE_GLOBAL_FUNC enclave_entry
[...]
.cfi_startproc

    /* Clear unused general registers */
    xor     %xdx, %xdx
    add     %xdx, %xdx          /* OF = SF = AF = CF = 0; ZF = PF = 1 */
    cld                         /* DF = 0 */
#if defined(LINUX64)
    xor     %r8, %r8
    xor     %r9, %r9
[...]

This routine somewhat finally invokes enter_enclave, which in turn dispatches enclave routines based on the leaf code.

I am struggling to know the meaning of the -e flag, and how does it come to play when transitioning into enclave.